Venn explanation:
Not looking at entire
sample space since
you know A must
occur – so you want
the likelihood of B
w.r.t the occurrence
of A
A
B
1
M = male, F= female, S = favors smoking, O = opposes smoking
** Good to write the English question as a math question **
a) You know you are randomly selecting a male, so “male” is the “given” and the
sample space is reduced to males
P(O | M) = 35 / 80 = P(O and M) / P(M) = (35/150) / (80/150)
b) You know you are randomly selecting a proponent of smoking, so “favors smoking”
is the “given” and the sample space is reduced to only those people
P(M | S) = 45 / 55 = P(M and S) / P(S) = (45/150) / (55/150)
2
“If it is known” implies the given:
Turn the sentence into a math sentence: P(2 | Sum = 4)
Which is equal to P(2 and Sum = 4) / P (Sum = 4)
Use the table to find these values:
11
12
13
14
15
16
21
22
23
24
25
26
31
32
33
34
35
36
41
42
43
44
45
46
51
52
53
54
55
56
61
62
63
64
65
66
P(22) / P(13, 22, 31) = (1/36) / (3/36) = 1/3
Then explain how you can just look at the reduced sample space:
Since you know the sum will be 4, then you are reduced to only looking at 3
outcomes, of which one contains a 2.
3
Let A be the event that the car lasts 5 yrs or more and
B be the event that the car lasts 10 yrs or more
A
B
Time=0
Time=5
Time=10
Therefore we know P(A)=0.95 and P(B)=0.45
If you look on the number line, you can see B is a proper subset of A, which means
that P( A and B) = P(B)= 0.45
We are asked to find P(B | A) = P(B and A) / P(A) = 0.45 / 0.95 = 9/19
= approx 0.4737
Explain how this is different than just determining the prob that a car will last 10 yrs
or more. -> If the car has made it to the 5 yr mark, it has a greater chance of making it
to the10 yr mark (0.4737 > 0.45)
4
a) P(A and B) / P(A) = ? / 0.8
b) P(A and B) / P(B) = ? / (1-0.4)
Use the fact that P(A un B) = P(A) + P(B) – P(A and B) to find ?
0.9
= 0.8 + 0.6 - ?
? = 1.4 – 0.9 = 0.5
So we have
a) 0.5 / 0.8 = 5/8 = 0.625
b) 0.5 / 0.6 = 5/6 = approx 0.833
Compare these values to P(B) and P(A), resp. to show how things changes with
additional knowledge
** Could use a Venn diagram **
5
Where are these derived from?
The conditional prob formula (look at last slide)!
Can be extended to more than 2 events but we will discuss this later using a different
technique. (trees)
**Stress this is the prob of one times the COND PROB of the other
(NOT P(A) * P(B))!!!!!!!!!!!!!!!!
6
P(A and B) = P(A) P( B | A) = (1-0.4)(0.5) = 0.3
**Could start from cond prob formula and cross multiply and not have to use the
product rule, per se.
7
8
a) Define events – mention you don’t want to use H to mean two different things
Explain what all the numbers mean in terms of probability -> cond. prob. Or not
Point out the sum of all prob branches from one node have a sum of 1
0.25
h
0.5
0.5
M
0.25 A
0.25
T
0.25
H
M
0.25
0.25 A
b) “If” indicates the given:0.25 T
0.25 on the tree)
P( H | h) = 0.25 (the branch
H
t
9
c) P( h and H) = P(h) * P(H|h) by the Product Rule
= 0.5 (0.25) = 1/8 (the path on the tree thru h and H)
d) P(H) is not on the tree (explain this)
P(H) = P[(h and H) or (t and H)]
= P(h and H) + P( t and H)
= 0.5(0.25) + (0.5)(0.25) (all the paths to get to H)
= 0.25
10
Ask students how tree should be started and how they know this from the problem.
a) F = fertilized lawn, U = unhealthy patches
0.10
F
70/
85
U
0.90
UC
15/85
FC
0.95
U
0.05
UC
11
Change into math sentences and read the tree:
b) P(U^C | F) = 0.9 (the branch on the tree)
c) P(F and U) = (70/85)(0.1) = 7/85 (the path on the tree thru F and U)
d) P(U) = (70/85)(0.1) + (15/85)(0.95) = ¼ (all the paths to get to U)
12
Talk about independent events
Ex (Ideas):
Indep -> The event I wear purple on any given day and the event that I
go out to eat
Dep -> The event that you bring an umbrella to school and the event that
you watch the weather channel before you leave your house
Different than mutually exclusive!!!!
Notice that P(A and B) = P(A)P(B) is different than the prod rule except when you
have independent events
13
a) (From Slides 15 and 16)
Check: P(H and h) ? P(H) P(h)
(0.5)(0.25) ? [0.25](0.5)
=
Independent
b) (From Slides 17 and 18)
Check: P(F and U) ? P(F) P(U)
(70/85)(0.1) ? (70/85) [1/4]
Not equal
Not Independent -> Dependent
14
All kids remember independently from one another.
a) P(K and Ke and M) = P(K)P(Ke)P(M) = 0.99(0.98)(0.975) = 0.945945
b) P(only one remembers) = P(K and not Ke and not M) +
P(not K and Ke and not M) +
P(not K and not Ke and M)
Since P(not K) = 1-0.99 = 0.01
P(not Ke) = 1-0.98 = 0.02
P(not M) = 1-0.975 = 0.025
We get (0.99)(0.02)(0.025) + (0.01)(0.98)(0.025) + (0.01)(0.02)(0.975)
= 0.000935
Can use a tree - ask them how this would be drawn.
15
Note: Will use this formula in its entirety more in stat class, but it is impt to notice
that the sample space must be completely partitioned (which will always be the case
in this class)
We will focus on this formula in a smaller case on the next slide.
16
Place P(A), P(B), P(E|A), etc on all the branches (ask students what should go where)
P(E|A) E
P(A)
A
P(EC |
A)
E
P(E | B) CE
B
P(EC | B)
P(B)
Bayes: P(A | E) =
P(A)P(E|A) /
P(A)P(E|A) + P(B)P(E|B)
= P(A and E) /
P(A and E) + P(B and E)
= P(A and E) / P(E)
(the same formula for cond prob)
Idea: You want the likelihood of taking
a certain path to end up where you
know you will end up - > find the prob
of the path you want and divide by
the prob of all possible paths to get to
your end pt.
EC
17
a) P(A) = 0.55, P(E |A) = 0.25, P(D | B) = 0.9, P(D|C) = 0.2
b) 0.25 (the branch on the tree) (*Could use the formula – but not necessary)
c) P(A and E) = (0.55)(0.25) = 0.1375 = 11/80 (the path thru A and E)
d) P(E) = P(A and E) +P(B and E) +P(C and E)
= (0.55)(0.25) + (0.3)(0.1) + (0.15)(0.8) = 0.2875 = 23/80
(all the paths to get to E)
e) P(A | E) = P(A and E) / P(E) (since this is not a branch on the tree)
= [0.1375] / [0.2875] (from above)
= 11/23
f) P(C|E) = P(C and E) / P(E) (since this is not a branch on the tree)
= (0.15)(0.8) / [0.2875] = 48/115
18
Ask students what to do to solve (where to begin the tree)
A,B,C = store where shirt is sold, S= shirt bought on sale
0.1 S
0.9NS
A
0.2
0.65 0.05 S
B
0.95
0.15 0.2 NS
P(S) = (0.2)(.1) + (0.65)(0.05) + (0.15)(0.2)
= 0.0825
S
P(A | S) = P(A and S) C
/ P(S) = 0.8
(0.2)(0.1) / [0.0825] = 8 / 33
P(B | S) = P(B and S) / P(S) = (0.65)(0.05)
NS / [0.0825] = 13 / 33
P(C | S) = P(C and S) / P(S) = (0.15)(0.2) / [0.0825] = 4 / 11 = 12/33
So the most likely store is Store B.
19
In succession -> key that a tree will be used.
Point out that you know the second card is a diamond, so on the tree you will have a “diamond” branch in the
second stage.
What will affect the likelihood of the second card being a diamond??
Taking out a diamond on the first – meaning that you will need a diamond branch in the 1st stage
You also need a branch allowing for a heart on the first since you are asked this in the question.
H,D,B = heart, diamond, black on first; d = diamond on the second
P(H | d) = P(H and d) / P(d)
= (13/52)(13/51) / [ (13/52)(13/51)+ (13/52)(12/51)+ (26/52)(13/51) ]
= 13 / 51
13/51 d
H
dC
12/51 d
13/52
13/52
D
26/52
dC
13/51 d
B
dC
20
Point out that you know the second card is a heart – what will affect the likelihood of the second card being a
heart??
(Taking out a heart on the first –> need a heart branch in the 1st stage)
Since this means you will need a branch allowing for a heart on the first, you can’t group all the reds together so
you will need a diamond branch as well to acct for all the different first red cards in the first stage.
H,D,B = heart, diamond, black on first; h = heart on the second
P(red | h) = P( H or D | h) = P( (H or D) and h ) / P(h)
= P( (H and h) or (D and h) ) / P(h) [all the paths you want / all possible paths to heart 2nd]
= [(13/52)(12/51) + (13/52)(13/51)] / [ (13/52)(12/51)+ (13/52)(13/51)+ (26/52)(13/51) ]
= 25 / 51
12/51 h
H
13/52
13/52 D
26/52
hC
13/51 h
hC
13/51 h
B
hC
21