Conditional
Probability,
Independent Events, Multiplicative
Rules, Baye's Rule.
Independent and dependent events:
Independent Events
When two events are said to be independent of each other, what this means is that
the probability that one event occurs in no way affects the probability of the other
event occurring. An example of two independent events is as follows; say you rolled a
die and flipped a coin. The probability of getting any number face on the die in no
way influences the probability of getting a head or a tail on the coin.
Dependent Events
When two events are said to be dependent, the probability of one event occurring
influences the likelihood of the other event.
For example, if you were to draw a two cards from a deck of 52 cards. If on your first
draw you had an ace and you put that aside, the probability of drawing an ace on the
second draw is greatly changed because you drew an ace the first time. Let's calculate
these different probabilities to see what's going on.
There are 4 Aces in a deck of 52 cards
On your first draw, the probability of getting an ace is given by:
If we don't return this card into the deck, the probability of drawing an ace on the
second pick is given by
As you can clearly see, the above two probabilities are different, so we say that the
two events are dependent. The likelihood of the second event depends on what
happens in the first event.
The Multiplication Rule:
Consider events A and B. P(A
B)= P(A)
P(B).
Note: Some books will say to take care that A and B are independent, but the rule can
also be used with dependent events, you just have to be more careful in find P(A) and
P(B).
What The Rule Means:
Suppose we roll one die followed by another and want to find the probability of
rolling a 4 on the first die and rolling an even number on the second die. Notice in this
problem we are not dealing with the sum of both dice. We are only dealing with the
probability of 4 on one die only and then, as a separate event, the probability of an
even number on one die only.
P(4)
P(even) = 3/6
=
1/6
So P(4 even) = (1/6)(3/6) = 3/36 = 1/12
While the rule can be applied regardless of dependence or independence of events, we
should note here that rolling a 4 on one die followed by rolling an even number on the
second die are independent events. Each die is treated as a separate thing and what
happens on the first die does not influence or effect what happens on the second die.
This is our basic definition of independent events: the outcome of one event does not
influence or effect the outcome of another event.
We’ll look at examples later that deal with dependent events. Just keep in mind that
what happens on one event will affect the other event.
Let's practice:
i.
Suppose you have a box with 3 blue marbles, 2 red marbles, and 4 yellow
marbles. You are going to pull out one marble, record its color, put it back in
the box and draw another marble. What is the probability of pulling out a red
marble followed by a blue marble?
The multiplication rule says we need to find P(red)
P(blue).
P(red) = 2/9.
P(blue)=3/9.
P(red blue) = (2/9)(3/9) = 6/81 = 2/27
The events in this example were independent. Once the first marble was pulled out
and its color recorded, it was returned to the box. Therefore, the probability for the
second marble was not effected by what happened on the first marble.
Notice that the final answer is always simplified. Some students find it helpful to
simplify before multiplying, but the final answer must always be simplified.
ii.
Consider the same box of marbles as in the previous example. However in
this case, we are going to pull out the first marble, leave it out, and then pull
out another marble. What is the probability of pulling out a red marble
followed by a blue marble?
We can still use the multiplication rule which says we need to find P(red) P(blue).
But be aware that in this case when we go to pull out the second marble, there will
only be 8 marbles left in the bag.
P(red)=2/9
P(blue)=3/8
P(red blue) = (2/9)(3/8) = 6/72 = 1/12
The events in this example were dependent. When the first marble was pulled out
and kept out, it effected the probability of the second event. This is what is meant by
dependent events.
iii.
Suppose you are going to draw two cards from a standard deck. What is
the probability that the first card is an ace and the second card is a jack (just
one of several ways to get “blackjack” or 21).
Using the multiplication rule we get
P(ace)
P(jack) = (4/52)(4/51) = 16/2652 = 4/663
Notice that this will be the same probability even if the question had asked for
the probability of a jack followed by an ace.
Conditional Probability:
The conditional probability of an event B is the probability that the event will occur
given the knowledge that an event A has already occurred. This probability is written
P(B|A), notation for the probability of B given A. In the case where events A and B are
independent (where event A has no effect on the probability of event B), the
conditional probability of event B given event A is simply the probability of event B,
that is P(B).
If events A and B are not independent, then the probability of the intersection of
A and B (the probability that both events occur) is defined by
P(A and B) = P(A)P(B|A).
From this definition, the conditional probability P(B|A) is easily obtained by dividing
by P(A):
P( B / A) 
P( A
and
B)
P( A)
or P( B / A) 
P( A  B )
P( A)
And also the conditional probability P ( A / B ) is obtained by dividing by P(B ).
P( A / B ) 
P( A
and
B)
or P( A / B ) 
P( A  B )
P( B )
P( B )
Examples:
Example1: In a card game, suppose a player needs to draw two cards of the same
suit in order to win. Of the 52 cards, there are 13 cards in each suit. Suppose first the
player draws a heart. Now the player wishes to draw a second heart. Since one heart
has already been chosen, there are now 12 hearts remaining in a deck of 51 cards. So
the conditional probability P(Draw second heart|First card a heart) = 12/51.
Example2: Suppose an individual applying to a college determines that he has an
80% chance of being accepted, and he knows that dormitory housing will only be
provided for 60% of all of the accepted students. The chance of the student being
accepted
and
receiving
dormitory
housing
is
defined
by
P(Accepted and Dormitory Housing) = P(Dormitory Housing|Accepted)P(Accepted)
= (0.60)*(0.80) = 0.48.
Example3: Suppose that somebody secretly rolls two fair six-sided dice, and we
must predict the outcome.


Let A be the value rolled on die 1
Let B be the value rolled on die 2
What is the probability that A = 2?
Table 1 shows the sample space of 36 outcomes
Clearly, A = 2 in exactly 6 of the 36 outcomes, thus P ( A  2) 
Table 1
+
A=1
B=1
2
2
3
3
4
4
5
5
6
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
6 1

36 6
Suppose it is revealed that A+B ≤ 5
What is the probability A+B ≤ 5 ?
Table 2 shows that A+B ≤ 5 for exactly 10 of the same 36 outcomes, thus
10 5
P( A  B  5) 

36 18
Table 2
+
A=1
B=1
2
2
3
3
4
4
5
5
6
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
What is the probability that A = 2 given that A+B ≤ 5 ?
Table 3 shows that for 3 of these 10 outcomes, A = 2
Thus, the conditional probability P( A  2 / A  B  5) 
Table 3
3
10
+
B=1
2
3
4
5
6
A=1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
Bayes's formula:
Another important method for calculating conditional probabilities is given by
Bayes's formula. The formula is based on the expression P(B) = P(B|A)P(A) +
P(B|Ac)P(Ac), which simply states that the probability of event B is the sum of the
conditional probabilities of event B given that event A has or has not occurred. For
independent events A and B, this is equal to P(B)P(A) + P(B)P(Ac) = P(B)(P(A) +
P(Ac)) = P(B)(1) = P(B), since the probability of an event and its complement must
always sum to 1. Bayes's formula is defined as follows:
P( B / A) P( A)
P( A / B ) 
P( B / A) P( A)  P( B / Ac ) P( Ac )