Section 6.3 – Conditional Probability
Pierced ears?
Male
19
71
90
Yes
No
Total
Female
84
4
88
Total
103
75
178
1. P(person has pierced ears given that they’re male) =
P(person has pierced ears given that they’re male)
Condition
# of Male who have pierced ears 19

Total # of males
90
Condition
2. P(person is male given that they have pierced ears) =
# of Males who have pierced ears
Total # of people who have pierced ears
=
19
103
Conditional Probability:
P( A  B)
 on your AP Formula Chart
P( B)
P(A | B) is read as the probability of “A given B”. We’re always given the condition.
For any events A & B, independent or dependent P( A | B) 
Let P = having pierced ears, M = being male.
P(M and P) = 19 / 178
P(M) = 90 / 178
19
P( P  M )
19 178 19
P( P | M ) 
 178 


90
P( M )
178 90 90
178
General Multiplication Rule
P( A  B)
P( B)  P( A | B) 
 P( B)
P( B)
P( B)  P( A | B)  P( A  B)
Example:
29% of Internet users download music. 67% of downloaders don’t care about copyright. What % of people
download AND don’t care?
Event D = Music Downloaders
P(D) = 0.29
P(C | D) = 0.67
Event C = Don’t Care
P(D and C) = P(C | D) · P(D)
= 0.67 · 0.29 = 0.1943
Example, p. 325
Suppose we know the following information about breast cancer and mammograms in a
particular region.
 1 % of the women age 40 or over in this region have breast cancer.
 For women who have breast cancer, the probability of a negative mammogram is 0.03.
 For women who don’t have breast cancer, the probability of a positive mammogram is 0.06.
A randomly selected woman aged 40 and over from this region tests positive for breast cancer in a
mammogram. Find the probability that she actually has breast cancer. Show your work.
Joint Probability
 P(A∩B)
Conditional Probability
 P(A|B)
Condition
 P(B)
Has breast cancer and
positive mammogram
0.01 * 0.97 = 0.0097
Has breast cancer and
negative mammogram
0.01 * 0.03 = 0.0003
Doesn’t have breast
cancer and positive
mammogram
0.99 * 0.06 = 0.0594
Doesn’t have breast
cancer and positive
mammogram
0.99 * 0.94 = 0.9306
We want 𝑃(ℎ𝑎𝑠 𝑏𝑟𝑒𝑎𝑠𝑡 𝑐𝑎𝑛𝑐𝑒𝑟|𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑚𝑎𝑚𝑚𝑜𝑔𝑟𝑎𝑚) =
𝑃(ℎ𝑎𝑠 𝑏𝑟𝑒𝑎𝑠𝑡 𝑐𝑎𝑛𝑐𝑒𝑟 𝑎𝑛𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑚𝑎𝑚𝑚𝑜𝑔𝑟𝑎𝑚)
=
𝑃(ℎ𝑎𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑚𝑎𝑚𝑚𝑜𝑔𝑟𝑎𝑚)
0.0097
0.0097+0.0594
=
0.0097
0.0691
.
≈ 0.14
If they ask for context: Given that a randomly selected woman from the region has a positive mammogram,
there is only about a 14% chance that she actually has breast cancer.
Finding this from a two-way table: (Suppose we have 10,000 women aged 40 and over)
Mammogram Positive
result
Negative
Total
Has Breast
Cancer?
Yes
No
97
594
3
9306
100
9900
P(B | P) = P(B ∩ P) / P(P) = 97 / 691 ≈ 0.14
Total
691
9309
10,000
INDEPENDENCE
Back to pierced ears example
P(P) =
103
 0.5787
178
P(P | M) =
19
 0.2111
90
P(P | F) =
84
 0.9545
88
If events are independent, the condition shouldn’t matter.
Since P(P) ≠ P(P | M) and P(P) ≠ P(P | F), this shows the condition matters.
NOT INDEPENDENT
Ex. Toss a coin twice.
P(1st tail) = 0.5
P(2nd tail) = 0.5
P(2nd tail | 1st tail) = 0.5
Coin flips are independent, so condition doesn’t change the original probability
*For independent events A & B
P(A | B) = P(A)  Condition doesn’t matter.
General Multiplication Rule
P(A & B) = P(A | B) · P(B)
If A & B are independent  = P(A) · P(B)
**Probability of 3 events intersecting:
P(A  B  C) = P(A) · P(B | A) · P(C | A and B)
All three happen
“B given A”
“C given A and B”
Example, p. 329
Assuming that O-ring joints succeed or fail independently, find the probability that the shuttle would launch
safely under similar conditions. P(O-ring succeeds) = 0.977
Because all six have to work: Letting P(JN) = Probability of that particular O-ring succeeding
P(J1 ∩ J2 ∩ J3 ∩ J4 ∩ J5 ∩ J6) = P(J1)* P(J2)* P(J3)* P(J4)* P(J5)* P(J6)
= (0.977) (0.977) (0.977) (0.977) (0.977) (0.977)
≈ 0.87
There is an 87% chance that the shuttle will launch safely under similar conditions.
***Important note: Events cannot be both mutually exclusive and independent!
HOMEWORK: p. 333 # 65, 67, 77, 85, 89, 91