Advanced Math Topics
5.5 Bayes’ Formula
A prisoner has escaped from jail. There are three roads leading away from jail.
If the prisoner selects road A, the probability of success is ¼. If he selects road B,
the probability of success is 1/5. If he selects road C, the probability of success is
1/6. The probability that the prisoner selects each of the roads is the same, 1/3.
If the prisoner succeeds in his escape, what is the probability that he made his
successful escape using road B?
Probability that you pick and are successful on each road…
(1/3)(1/4)
A
(1/3)(1/5)
B
(1/3)(1/6)
C
We can come up with Bayes’ formula using these expressions.
P(picked B | success) =
favorable probability of success
total probability of success
p(picked B) • p(success | picked B)
=
p(picked A) • p(success | picked A) + p(picked B) • p(success | picked B) + p(picked C) • p(success | picked C)
=
1/3 • 1/5
1/3 • 1/4 + 1/3 • 1/5 + 1/3 • 1/6
=
1/15
1/12 + 1/15 + 1/18
=
0.06667
0.205556
= 32.43%
Given that the prisoner was succesful, there is a 32.43% probability he used road B.
P(picked B | success) =
p(picked B) • p(success | picked B)
p(picked A) • p(success | picked A) + p(picked B) • p(success | picked B) + p(picked C) • p(success | picked C)
Bayes’ Formula
A sample space of mutually exclusive events A1, A2, …An gives the following.
P(A1 | B) =
p(A1) • p(B | A1)
p(A1) • p(B | A1) + p(A2) • p(B | A2) + …..+ p(An) • p(B | An)
From the HW P. 275
1) Truck 1 has 6 defective and 32 good microwaves. Truck 2 has 5 defective
and 16 good microwaves. An inspector randomly selects a microwave and finds
that it is defective. He forgot which truck it came from. If there is a ½ probability that
it came from either truck, find the probability it came from truck 2.
(1/2)(5/21)
(1/2)(5/21) + (1/2)(6/38)
= 60.13%
From the HW P. 275
3) A hospital receives its blood supply from four banks: A, B, C, and D. The probabilities
that the blood received from each bank is contaminated is 0.11, 0.09, 0.13, and 0.10,
respectively. The hospital randomly selects a blood bank. If the blood received is
contaminated, what is the probability that the blood is from Bank B?
(1/4)(0.09)
= 20.93%
(1/4)(.11) + (1/4)(.09) + (1/4)(.13) + (1/4)(.10)
HW
P. 275 #1-6,8