Download In-class Exercises on Probability, 24 October 2011 Name. Hints and

In-class Exercises on Probability, 24 October 2011
Name.
Hints and Answers
1. An assignment contains 20 questions.
(a) In how many ways can the grader choose 10 of those 20 questions to grade?
(b) Suppose Winton did 17 of the 20 questions. What is the probability that Winton did all
of the 10 questions the grader will randomly choose to grade?
(c) There are 40 runners in the final of the olympic marathon. In how many ways can gold,
silver and bronze medals be awarded to the runners?
20!
= 184, 756.
10!10!
(b) The number of groups of 10 problems Winton did is C17,10 = 19, 448. Therefore, the
probability that the reader will choose one of Winton’s 19,448 groups of 10 out of the total
possible 184,756 groups of 10 is:
19, 448
= .1053
184, 756
Answer. (a) C20,10 =
(c) From basic counting there are 40 ways to award the gold medal, once that has been done,
there are 39 possible runners to win the silver, and then there will be 38 possible choices
of runners for the silver medal. By the multiplication rule, there are 40 · 39 · 38 = 59, 280
possible ways to award the medals to the runners. (Alternatively, this is a problem that
40!
= 59, 280
involves permuting 40 objects taken 3 at a time, thus there are P40,3 =
(40 − 3)!
possibilities.)
2. Labron James is a 75% free throw shooter. Assume the success of one free throw to another
is independent. Suppose he attempts 15 free throws in one game. Use the binomial distribution
to answer the following questions.
(a) What is probability that he will make exactly 12 of the 15 free throws?
(b) What is the probability that he will make all of the 15 free throws?
(c) What is the probability that he will make at least 12 of the 15 free throws? (Feel free to
use Table 3 in Appendix II for this part).
Answer. Use the binomial distribution with n = 15, p = .75 and q = .25.
(a) P (r = 12) = C15,12 (.75)12 (.25)3 =
Wolfram Alpha)
15!
(.75)12 (.25)3 ≈ .2252 (Compare with the answer at
12!3!
(b) P (r = 15) = (.75)15 ≈ .01336
(c) With the help of the table:
P (r ≥ 12) = P (r = 12) + P (r = 13) + P (r = 14) + P (r = 15) ≈ .225 + .156 + .067 + .013 = .461
Compare with the answer at this Wolfram Alpha link
For further examples and motivation of the binomial distribution, see the Kahn Academy video
Introduction to the Binomial Distribution and the three videos that follow it.
3. (From Text) Sara is a 60-year old female in reasonably good health. She wants to take out
a $50,000 term life insurance policy until she is 65 (the policy expires on her 65th birthday).
The probability of death in a given year for women in Sara’s demographic are:
x = age
60
P(death at this age) 0.00756
61
0.00825
62
0.00896
63
0.00965
64
0.01035
Sara is applying to Big Rock Insurance Company for her term policy.
(a) What is the probability that Sara will die in her 60th year. Using this probability and the
$50,000 death benefit, was is the expected cost to Big Rock Insurance?
(b) Repeat part (a) for years 61,62,63,64. What would the total expected cost to Big Rock
Insurance over the years 60 through 64. How does this relate to the expected value of a random
variable?
(c) If Big Rock Insurance wants to make a profit of $700 above the expected total cost paid
out for Sara’s death, how much should it charge for the policy?
(d) If Big Rock charges $5000 for the policy, how much profit does the company expect to
make?
Answer. (a) P (60), means, e.g., the probability Sara will die in her 60th year; so from the
table P (60) = .0756, and the expected cost is (.0756)(50000) = $378.
(b) P (61) = .00825, Expected Cost is (.00825)(50000) = $412.5.
P (62) = .00896, Expected Cost is (.00896)(50000) = $448.
P (63) = .00965, Expected Cost is (.00965)(50000) = $482.5.
P (64) = .01035, Expected Cost is (.01035)(50000) = $517.5.
(c) 378 + 412.5 + 448 + 482.5 + 517.5 + 700 = $2938.50
(d) The expected profit is (assuming no company overhead): $5000. − $2238.50 = $2761.50