Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 5 October 15, 2009

3.39. (a) Y = the number of components that last longer than 1000 hours has a binomial distribution. P (Y = 2) = \binom{4}{2}(.8)^{2}(.2)^{2} ≐ .1536.

(b) The subsystem will operate if 2, 3, or 4 of the components last longer than 1000 hours. So this probability is P (Y ≥ 2) = \binom{4}{2}(.8)^{2}(.2)^{2} + \binom{4}{3}(.8)^{3}(.2) + \binom{4}{4}(.8)^{4} ≐ .973.

3.40. These are all binomial probabilities. Let Y be the number who recover out of the 20. (a) P (Y = 14) = \binom{14}{20}(.8)^{14}(.2)^{6} ≐ .11.

(b) P (Y ≥ 10) = ∑_{y=10}^{20} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .9994.

(c) P (14 ≤ Y ≤ 18) = ∑_{y=14}^{18} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .844.

(d) P (0 ≤ Y ≤ 16) = ∑_{y=0}^{16} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .589.

3.43. Let Y be the number who qualify for the favorable rates (out of the 5), a binomial random variable. (a) P (Y = 5) = (.7)^{5} ≐ .168.

(b) P (Y ≥ 4) = ∑_{y=4}^{5} \binom{5}{y}(.7)^{y}(.3)^{5−y} ≐ .528.

3.44. Let Y be the number of successful operations (binomial). (a) P (Y = 5) = (.8)^{5} ≐ .328. (b) P (Y = 4) = \binom{5}{4}(.6)^{4}(.4) ≐ .259.

(c) P (Y < 2) = ∑_{y=0}^{1} \binom{5}{y}(.3)^{y}(.7)^{5−y} ≐ .528. (Note that this is the same as the answer for 3.43

(b). Do you see why? For a formal statement of the pattern here, see Exercise 3.54.)

3.53. Y = the number of children (out of 3) that develop Tay-Sachs has a binomial distribution.

(a) P (Y = 3) = (.25)^{3} ≐ .0156.

(b) P (Y = 1) = \binom{3}{1}(.25)(.75)^{2} ≐ .422.

(c) This conditional probability is \frac{(.75)^{2}(.25)}{(.75)^{2}} = .25. (This follows because of the independence.)

3.59. Let Y be the number of good motors, so 10 − Y is the number of defectives. The problem is asking for the expected value of G = 1000 − 200(10 − Y) = 200Y − 1000. Since Y is binomial with n = 10 and p = .92, we have E(G) = E(200Y −10000) = 200E(Y)−1000 = (200)(9.2)−1000 = 840 (dollars).

3.62. (a) To justify multiplying p_{1}p_{2}p_{3} to get the probability of detecting a wing crack, you need to assume the individual events are independent.

(b) p = (.9)(.8)(.5) = .36. Then the probability desired is P (Y ≥ 1) = ∑_{y=1}^{3} \binom{3}{y}(.36)^{y}(.64)^{3−y} ≐ .738. 3.67. This is geometric with p = .3. So P (Y = 5) = (.7)^{4}(.3) ≐ .072. 3.69. Y is a geometric random variable with p = .59, so P (Y = y) = (.41)^{y−1}(.59).