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MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cards in the deck but we are “given that”
the card is not a face card.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cards in the deck but we are “given that”
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cards in the deck but we are “given that”
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
So the conditional sample space consists of the
52 – 12= 40 cards that are NOT face cards.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cardsThere
in theare
deck
but we are “given that”
4 twos.
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
So the conditional sample space consists of the
52 – 12= 40 cards that are NOT face cards.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cardsThere
in theare
deck
but we are “given that”
4 twos.
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the…and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 – 12= 40 cards that are NOT face cards.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cardsThere
in theare
deck
but we are “given that”
4 twos.
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the…and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 – 12= 40 cards that are NOT face cards.
4
1
𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cardsThere
in theare
deck
but we are “given that”
4 twos.
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the…and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 – 12= 40 cards that are NOT face cards.
4
1
𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cardsThere
in theare
deck
but we are “given that”
4 twos.
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the…and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 – 12= 40 cards that are NOT face cards.
4
1
𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find 𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘).
There are 52 cards in the deck but we are “given that”
the card is not a face card.
That means that the “given that” condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
So the conditional sample space consists of the
52 – 12= 40 cards that are NOT face cards.
4
1
𝑃 π‘‘π‘€π‘œ π‘›π‘œπ‘›π‘“π‘Žπ‘π‘’ π‘π‘Žπ‘Ÿπ‘‘) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
On the first draw, the
card could be a diamond
or not a diamond.
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
On the first draw, the
card could be a diamond
or not a diamond.
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume
2 cards are
There arethat
13 diamonds
& drawn from a standard 52-card deck.
If the cards are
drawn without replacement, find the
39 non-diamonds
in the
probability
of drawing
52 card deck
giving us a diamond followed by a non-diamond.
thesemeans
probabilities
That
that we need to compute this probability:
Diamond
1st
draw
NonDiamond
On the first draw, the
card could be a diamond
or not a diamond.
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st draw
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
Both of these
fractions can
be simplified.
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st draw
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
don’tfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
probability of drawing a diamond
followed
‘1st is Diamond’
and by
‘2ndaisnon-diamond.
not Diamond’
branch.
That means that we need to compute
this probability:
Diamond
1st draw
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
don’tfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
probability of drawing a diamond
followed
‘1st is Diamond’
and by
‘2ndaisnon-diamond.
not Diamond’
branch.
That means that we need to compute
this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
don’tfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
probability of drawing a diamond
followed
‘1st is Diamond’
and by
‘2ndaisnon-diamond.
not Diamond’
branch.
That means that we need to compute
this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
We are drawing without
replacement so after the
1st card was drawn
(which on this branch
was a diamond), there
are only 51 cards left (12
diamonds and 39 nondiamonds).
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
don’tfrom
needathe
entire tree
diagram.
theprobability
cards areofdrawn without
All replacement,
we really need isfind
the the
So,Ifthe
st is Diamond’
ndaisnon-diamond.
probability
of
drawing
a
diamond
followed
by
‘1
and
‘2
not Diamond’
drawing a nonbranch.
That
means
we
need
to
compute
this probability:
diamond
is 39 that
.
51
Diamond
1st
draw
NonDiamond
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
We are drawing without
replacement so after the
1st card was drawn
(which on this branch
was a diamond), there
are only 51 cards left (12
diamonds and 39 nondiamonds).
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
don’tfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
…whichof
reduces
probability
drawing a diamond
followed
‘1st is Diamond’
and by
‘2ndaisnon-diamond.
not Diamond’
13
to
17. that we need to compute
branch.
That means
this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
We are drawing without
replacement so after the
1st card was drawn
(which on this branch
was a diamond), there
are only 51 cards left (12
diamonds and 39 nondiamonds).
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
without drawing the tree diagram at all.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 14
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 14
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 • 13 17
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 • 13 17 =
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 • 13 17 = 13 68
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIf𝑖𝑠the
are
without
find
the
1𝑠𝑑
π‘Ž cards
1𝑠𝑑 𝑖𝑠 replacement,
π‘Žreplacement,
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑
𝑖𝑠drawn
π‘›π‘œπ‘‘
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
the
cards
are
drawn
without
find
𝑃
π‘Žπ‘›π‘‘
=𝑃
βˆ™π‘ƒ
| the
probability
followed
aanon-diamond.
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘ of
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘aadiamond
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 • 13 17 = 13 68
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1𝑠𝑑 𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠 π‘›π‘œπ‘‘
𝑃
π‘Žπ‘›π‘‘
= βˆ™
=
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
4 17 68
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
𝑖𝑠 π‘›π‘œπ‘‘ 1𝑠𝑑
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=𝑃
βˆ™ 𝑃 them to what
| we need
here.
We see that only
the probabilityπ‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
of getting a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
39
13
deck making the probability of
a• non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
Non-Diamond
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
𝑖𝑠 π‘›π‘œπ‘‘ 1𝑠𝑑
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=𝑃
βˆ™ 𝑃 them to what
| we need
here.
We see that only
the probabilityπ‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
of getting a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
39
13
deck making the probability of
a• non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
Non-Diamond
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
𝑖𝑠 π‘›π‘œπ‘‘ 1𝑠𝑑
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=𝑃
βˆ™ 𝑃 them to what
| we need
here.
We see that only
the probabilityπ‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
of getting a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
draw changes. Because the 1st card was replaced,39
there are39
still 52 cards 39
in the
52
1
39
13
deck making the probability of
a• non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
39
Non-Diamond
52
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
1𝑠𝑑
π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
them𝑖𝑠toπ‘›π‘œπ‘‘
what
we 𝑖𝑠
need
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=
𝑃
βˆ™
𝑃
|
39 probability
here.
We see that only
the
of3getting
a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
But
to
.
4
52 reduces
draw changes. Because the 1st card was replaced,39
there are39
still 52 cards 39
in the
52
1
39
13
deck making the probability of
a• non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
39
Non-Diamond
52
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
1𝑠𝑑
π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
them𝑖𝑠toπ‘›π‘œπ‘‘
what
we 𝑖𝑠
need
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=
𝑃
βˆ™
𝑃
|
39 probability
here.
We see that only
the
of3getting
a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
But
to
.
4
52 reduces
3
draw changes. Because the 1st card was replaced, there
are39
still 52 cards 39
in the
4
1
39
13
deck making the probability of
a• non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
3
Non-Diamond
4
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
𝑖𝑠 π‘›π‘œπ‘‘ 1𝑠𝑑
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=𝑃
βˆ™ 𝑃 them to what
| we need
here.
We see that only
the probabilityπ‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
of getting a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
3
deck making the probability of
a• non-diamond
instead of .
= drawing
=
52
51
4
4
Diamond
Non-Diamond
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
𝑖𝑠 π‘›π‘œπ‘‘ 1𝑠𝑑
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=𝑃
βˆ™ 𝑃 them to what
| we need
here.
We see that only
the probabilityπ‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
of getting a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
3
deck making the probability of
a• non-diamond
3instead of 51.
= drawing
=
52
4
4
16
Diamond
Non-Diamond
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Let’s examine the previous calculation and compare it to what we need to
So,they
the probability
of drawing
a
calculate this new probability. We see that
are exactly the
same EXCEPT,
because the 1st card is replaced before
the 2ndfollowed
draw, now
probability of
diamond
bythe
a39non-diamond
39
drawing a non-diamond nextwith
is replacement
instead of .is 3 16.
52
51
1𝑠𝑑 𝑖𝑠 π‘Ž
1𝑠𝑑and
𝑖𝑠 π‘Žcompare 2𝑛𝑑
𝑖𝑠 π‘Ž
2𝑛𝑑 𝑖𝑠done
π‘›π‘œπ‘‘ previously
𝑖𝑠 π‘›π‘œπ‘‘ 1𝑠𝑑
𝑃Look at the calculations
π‘Žπ‘›π‘‘
=𝑃
βˆ™ 𝑃 them to what
| we need
here.
We see that only
the probabilityπ‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
of getting a non-diamond
on π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
the second
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
π‘‘π‘–π‘Žπ‘šπ‘œπ‘›π‘‘
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
3
deck making the probability of
a• non-diamond
3instead of 51.
= drawing
=
52
4
4
16
Diamond
Non-Diamond
3
16
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find 𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜).
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find 𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜).
Here we are “given that” the disk is pink.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find 𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜).
X X X
X X X X
Here we are “given that” the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find 𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜).
X X X
X X X X
Here we are “given that” the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find 𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜).
X X X
X X X X
Here we are “given that” the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
1 1
𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜) = =
3 5
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find 𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜).
X X X
X X X X
Here we are “given that” the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
1 1
𝑃 π‘ π‘‘π‘Žπ‘Ÿ π‘π‘–π‘›π‘˜) = =
3 5
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
The hint reminds us that we can get a red and a green either by getting the
green first and the red second or by getting the red first and the green second.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
The hint reminds us that we can get a red and a green either by getting the
green first and the red second or by getting the red first and the green second.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
The hint reminds us that we can get a red and a green either by getting the
green first and the red second or by getting the red first and the green second.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
thedraw
first draw,
the probability that youOnwill
a
we ball?
can get either
red ball and a green
green, blue or red.
G
B
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
This reduces.
8
3
22
G
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
This reduces.
4
3
11
G
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110
Sec
13.3areConditional
Practice Exercises
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
G
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
On
the second4
the probability that you will draw
a
draw, we can still11 B
red ball and a green ball? get either green,
blue or red. 3
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
G
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
On
the second4
the probability that you will draw
a
draw, we can still11 B
red ball and a green ball? get either green,
blue or red. 3
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
G
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
On
the second4
the probability that you will draw
a
draw, we can still11 B
red ball and a green ball? get either green,
blue or red. 3
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls
red
What
Noticeand
that3we
onlyballs.
need the
2 is
branches containing
bothwill
R and
G. a
the probability
that you
draw
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls
red
What
Noticeand
that3we
onlyballs.
need the
2 is
branches containing
bothwill
R and
G. a
the probability
that you
draw
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice
Exercises
Now
there
only 21 ballsProbability
left. Find the
probability
of drawing
G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
red after drawing green.
G
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice
Exercises
Now
there
only 21 ballsProbability
left. Find the
probability
of drawing
G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
red after drawing green.
G
3
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
21
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
3
This reduces.
4
11
B
3
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
21
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
This reduces.
4
11
B
3
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
7
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
7
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
1
7 R
G
4
11
Find the probability
of drawing
B
a green after drawing red.B
(Hint: There are 2 ways this can happen.)
3
R
22
G
Let’s use a tree diagram to
calculate this probability.
R
B
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
1
7 R
G
4
11
Find the probability
of drawing
B
a green after drawing red.B
(Hint: There are 2 ways this can happen.)
3
R
22
G
Let’s use a tree diagram to
calculate this probability.
R
B
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
7
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1 11
3 11
2
βˆ™
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
βˆ™
=
2 14
7
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
3 11
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
βˆ™
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
3 11
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
βˆ™
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
3 11
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
βˆ™
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
3 11
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
βˆ™
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
1
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
1
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
1
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Let’s use a tree diagram to
calculate this probability.
G
1
4
3
11
7
B
22
R
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 = 𝑃 1𝑠𝑑 𝐺 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝑅 + 𝑃(1𝑠𝑑 𝑅 π‘Žπ‘›π‘‘ 2𝑛𝑑 𝐺)
1
1
2
𝑃 𝑅 π‘Žπ‘›π‘‘ 𝐺 =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st roll…so the rolls are independent.
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st roll…so the rolls are independent.
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™ =
1
6
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
and 𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™ =
16
26
36
46
56
66
1
6
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st roll…so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™ = and 𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™ =
6
6
1 1
1
π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘›
π‘ π‘’π‘š 𝑖𝑠 7
π‘ π‘’π‘š 𝑖𝑠 7
𝑃
=𝑃
βˆ™π‘ƒ
= βˆ™ =
6 6
36
π‘π‘œπ‘‘β„Ž π‘Ÿπ‘œπ‘™π‘™π‘ 
π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™
π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st roll…so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™ = and 𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™ =
6
6
1 1
1
π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘›
π‘ π‘’π‘š 𝑖𝑠 7
π‘ π‘’π‘š 𝑖𝑠 7
𝑃
=𝑃
βˆ™π‘ƒ
= βˆ™ =
6 6
36
π‘π‘œπ‘‘β„Ž π‘Ÿπ‘œπ‘™π‘™π‘ 
π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™
π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st roll…so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™ = and 𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™ =
6
6
1 1
1
π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘›
π‘ π‘’π‘š 𝑖𝑠 7
π‘ π‘’π‘š 𝑖𝑠 7
𝑃
=𝑃
βˆ™π‘ƒ
= βˆ™ =
6 6
36
π‘π‘œπ‘‘β„Ž π‘Ÿπ‘œπ‘™π‘™π‘ 
π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™
π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st roll…so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™ = and 𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™ =
6
6
1 1
1
π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘›
π‘ π‘’π‘š 𝑖𝑠 7
π‘ π‘’π‘š 𝑖𝑠 7
𝑃
=𝑃
βˆ™π‘ƒ
= βˆ™ =
6 6
36
π‘π‘œπ‘‘β„Ž π‘Ÿπ‘œπ‘™π‘™π‘ 
π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™
π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Let’s examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st roll…so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™ = and 𝑃 π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™ =
6
6
1 1
1
π‘ π‘’π‘š 𝑖𝑠 7 π‘œπ‘›
π‘ π‘’π‘š 𝑖𝑠 7
π‘ π‘’π‘š 𝑖𝑠 7
𝑃
=𝑃
βˆ™π‘ƒ
= βˆ™ =
6 6
36
π‘π‘œπ‘‘β„Ž π‘Ÿπ‘œπ‘™π‘™π‘ 
π‘œπ‘› 1𝑠𝑑 π‘Ÿπ‘œπ‘™π‘™
π‘œπ‘› 2𝑛𝑑 π‘Ÿπ‘œπ‘™π‘™
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q ∩ R),
associated with the tree diagram.
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q ∩ R),
associated with the tree diagram.
𝑃 𝑄 ∩ 𝑅 = 𝑃(𝑄 π‘Žπ‘›π‘‘ 𝑅)
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q ∩ R),
associated with the tree diagram.
𝑃 𝑄 ∩ 𝑅 = 𝑃(𝑄 π‘Žπ‘›π‘‘ 𝑅)
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q ∩ R),
associated with the tree diagram.
𝑃 𝑄 ∩ 𝑅 = 𝑃(𝑄 π‘Žπ‘›π‘‘ 𝑅)
𝑃 𝑄 ∩ 𝑅 = (0.4)(0.8)
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q ∩ R),
associated with the tree diagram.
𝑃 𝑄 ∩ 𝑅 = 𝑃(𝑄 π‘Žπ‘›π‘‘ 𝑅)
𝑃 𝑄 ∩ 𝑅 = (0.4)(0.8)
R
M
0.4 N
S
R
S
𝑃 𝑄 ∩ 𝑅 = 0.32
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
It is clear that the outcome of the penny flip has no effect on the
probability of any outcome of the nickel flip.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
It is clear that the outcome of the penny flip has no effect on the
probability of any outcome of the nickel flip.
Events A and B are independent.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
Be careful! You are only flipping each coin once and both events
mention only the penny. If the penny comes up heads (and you are just
tossing it once), then the probability the it comes up tails is 0.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
Be careful! You are only flipping each coin once and both events
mention only the penny. If the penny comes up heads (and you are just
tossing it once), then the probability the it comes up tails is 0.
Events A and B are dependent.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
𝑃 π‘ π‘’π‘Ÿπ‘£π‘–π‘π‘’ π‘”π‘œπ‘œπ‘‘ π‘‘π‘–π‘›π‘›π‘’π‘Ÿ) =
59
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
𝑃 π‘ π‘’π‘Ÿπ‘£π‘–π‘π‘’ π‘”π‘œπ‘œπ‘‘ π‘‘π‘–π‘›π‘›π‘’π‘Ÿ) =
59
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
𝑃 π‘ π‘’π‘Ÿπ‘£π‘–π‘π‘’ π‘”π‘œπ‘œπ‘‘ π‘‘π‘–π‘›π‘›π‘’π‘Ÿ) =
59
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
𝑃 π‘ π‘’π‘Ÿπ‘£π‘–π‘π‘’ π‘”π‘œπ‘œπ‘‘ π‘‘π‘–π‘›π‘›π‘’π‘Ÿ) =
59
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