Bowdoin Math 1200, Spring ’17 – Quiz #3 – Wednesday February 22, 2017
Time allotted: 15 minutes. No books, notes, homework, or “cheat sheets”, are allowed. You may use a calculator.
For full credit indicate your reasoning, show all of your work and write LEGIBLY. Do your work in the space provided
below and on the back of the sheet. Raise your hand if you have a question or need more paper.
Your Name (LAST, First): SOLUTIONS
Problem 1. (4 points) Two fair dice are rolled. What is the conditional probability that at least one lands on
a 6, given that the two dice land on different numbers? Hint: Just use the definition of conditional probability.
Problem 2. (6 points) People can be divided into two classes: those who are accident-prone and those who are
not. An insurance company’s statistics show that an accident-prone person will have an accident at some time
within a fixed 1-year period with probability 0.4, whereas this probability decreases to 0.2 for a person who is
not accident prone. Assume that 30% of the population is accident-prone.
(a) What is the probability that a new policyholder will have an accident within a year of purchasing a policy?
(b) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the
probability that he or she is accident-prone? (You may leave the answers in terms of fractions.)
Answers:
P1. We events A = “at least one die lands on a 6” and B = “the two dice land on different numbers”, in S:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
A
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }.
Note that the outcomes in B are all the off-diagonal elements of S, namely the outcomes not of the type
(i, i), with i = 1, 2, . . . , 6. Therefore #(B) = 62 − 6 = 30, and P (B) = #(B)/#(S) = 30/36 = 5/6. Also,
#(A) = 11, and the outcomes in A ∩ B are all those in A except (6, 6). Therefore #(A ∩ B) = 11 − 1 = 10
10/36
and P (A ∩ B) = #(A ∩ B)/#(S) = 10/36. In conclusion, P (A|B) = P (A ∩ B)/P (B) = 30/36
= 1/3 . P2. We define the events: A = “the policy holder is accident prone”, and X = “the policy holder has an accident
within a year”. With the data provided by the problem, the corresponding tree diagram is the following:
(a) By the Total Probability Theorem:
P (X) = P (X|A)P (A) + P (X|Ac )P (Ac ) =
(b) By Bayes’ rule: P (A|X) =
4 3
2 7
26
·
+
·
=
= 26% .
10 10 10 10
100
P (X|A)P (A)
=
P (X)
4
10
·
3
10
26
100
=
6
' 0.462 .
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