Mathematics 375 – Probability Theory Solutions for Problem Set 4 September 30, 2011

3.12. We have E(Y) = (1)(.4) + (2)(.3) + (3)(.2) + (4)(.1) = 2. Similarly E(1/Y) ≐ (1)(.4) + (.5)(.3) + (.333)(.2) + (.25)(.1) ≐ .642. Then E(Y^{2} − 1) = (0)(.4) + (3)(.3) + (8)(.2) + (15)(.1) = 4. Finally V (Y) = E(Y^{2}) − (E(Y))^{2} = 5 − 2^{2} = 1.

3.14 (a) E(Y) = 7.9. (b) V (Y) = E(Y^{2})−(E(Y))^{2} = 67.14−(7.9)^{2} = 4.73, so σ(Y) = \sqrt{4.73} ≐ 2.17.

(c) The interval μ ± 2σ is 7.9 ± 4.35 = [3.55, 12.25]. The probability that the patent life is in this range is 1 − .03 − .01 = .96. (This is quite good agreement with the empirical rule, of course.)

3.17. Recall from PS 3 that the probability mass function for Y = number of empty bowls was

P (Y = 0) = \frac{6}{27} = \frac{2}{9}
P (Y = 1) = \frac{18}{27} = \frac{2}{3}
P (Y = 2) = \frac{3}{27} = \frac{1}{9}.

So E(Y) = (0)(2/9)+(1)(2/3)+2(1/9) = 8/9 and V (Y) = E(Y^{2})−E(Y)^{2} = 10/9−64/81 = 26/81 and σ ≐ .567. The interval μ±2σ is approximately [−.24, 2.02]. All values of Y are in this interval, so the probability = 1.

3.18. Recall from PS 3 that the probability function for Y = number of cells in second generation was

P (Y = 0) = .1 + (.9)(.1)^{2} = .109
P (Y = 2) = (.9)((.1)(.9) + (.9)(.1)) = .162
P (Y = 4) = (.9)^{3} = .729.

So the mean number of cells in generation 2 is (0)(.109) + (2)(.162) + (4)(.729) = 3.24.

3.19. If C is the premium charged for the policy, then the company’s profit is C − 15 if the event A does not occur, and C − 15 − 1000 if it does. Since the probability of A is .02 and the probability of A not occurring is .98, the expected profit is E(P) = (.98)(C − 15) + (.02)(C − 15 − 1000). To make this equal 50, C should solve

50 = C − 15 − 20, so P = 85.

(Note that we need to use the probabilities .98 and .02 here, not the percentages.)

3.31. (a) Whether or not the mean of W = 2Y is larger than the mean of Y will depend on whether μ = E(Y) is positive or negative. If E(Y) > 0, then the mean of W will be larger; if E(Y) < 0, then it will be smaller. (b) By Theorem 3.4, E(W) = 2E(Y). This does agree with (a).

(c) The variance of W should also be larger since the values of W will be more spread out. (d) V (W) = E(W^{2}) − (E(W))^{2} = E(4Y^{2}) − (2E(Y))^{2}. This equals 4(E(Y^{2}) − (E(Y))^{2}) = 4V (Y).

3.33. (a) By Theorem 3.5, E(aY + b) = E(aY) + E(b). Then since a, b are constant, Theorems 3.4 and 3.3 imply this equals aE(Y) + b. (b) We have by part (a) and Theorems 3.3 - 3.5:

V (aY + b) = E((aY + b)^{2}) − (E(aY + b))^{2}
= E(a^{2}Y^{2} + 2abY + b^{2} − (aE(Y) + b)^{2}
= a^{2}E(Y^{2}) + 2abE(Y) + b^{2} − a^{2}(E(Y))^{2} − 2abE(Y) − b^{2}
= a^{2}(E(Y^{2}) − (E(Y))^{2})
= a^{2}V (Y).

3.37. Of these quantities, (c) and (e) could be approximated by binomial random variables. n = 100 for both, but p cannot be determined in either case from the given information. Note that (a) is not a random variable at all since the description does not say “in the sample.” The (total) number of students who took the SAT that year is a constant. (b) and (d) are random variables but they do not match the number of successes in a binomial experiments since there are not just two possible outcomes for the score or the time taken for each of the students in the sample.

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. (a) P (Y = 8) = \binom{20}{14}(.8)^{14}(.2)^{6} ≐ .109.

(b)

P (Y ≥ 10) = ∑_{y=10}^{20} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .999.

(c)

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

(d)

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

3.41. The probability is

P (Y ≥ 10) = ∑_{y=10}^{15} \binom{15}{y}(.2)^{y}(.8)^{15−y} ≐ .0001.

3.42. (a) If one of the incorrect answers can be eliminated each time, then the probability of finding the correct one by guessing goes up to p = .25. So the probability of getting at least 10 correct answers is now

P (Y ≥ 10) = ∑_{y=10}^{15} \binom{15}{y}(.25)^{y}(.75)^{15−y} ≐ .0008.

Similarly if two incorrect answers can be eliminated each time,

P (Y ≥ 10) = ∑_{y=10}^{15} \binom{15}{y}(.333)^{y}(.667)^{15−y} ≐ .008.

(In other words, this much-discussed “test-taking strategy” is really useful only when you can use it sparingly. Really knowing the material–so guessing is unnecessary most of the time–is a much superior strategy!!)

3.46. The probability histogram for p = .1 should have boxes of heights

.590490, .32805, 0.07290, 0.00810, 0.00045, 0.000010

(the first box extends from −1/2 to 1/2, etc.. This one is skewed to the left. With p = .5, the boxes have heights

0.03125, .15625, .3125, .3125, .15625, 0.03125

This one is symmetrical. Finally with p = .9 the heights are the same as for .1 but in reverse order. This one is skewed right.

3.48. (a) The number of sets that detect the incoming missile is binomial with n = 5 and p = .9. The probability that exactly 4 sets detect it is \binom{5}{4}(.9)^{4}(.1) = .32805. The probability that at least one detects it is

1 − P (Y = 0) = 1 − (.00001) = .99999.

(b) “Detecting” should mean that at least one set detects the incoming missile. The result of part (a) shows that n = 5 is certainly big enough. But in fact if n = 3,

1 − P (Y = 0) = 1 − \binom{3}{0}(.1)^{3} = .999.

So n = 3 is large enough.

3.64. We have P = P (Y = y_{0}) = \binom{n}{y_{0}}p^{y_{0}}(1 − p)^{n−y_{0}}, where we think of this as a function of the unknown value p. Given that y_{0} successes were observed, the question is to determine a best estimate for p.

This function of p will be maximized at a point where \frac{dP}{dp} = 0. In office hours, I suggested maximizing ln(P) instead because that simplifies the algebra. It is also possible to differentiate P directly, of course. We have by the product rule:

\frac{dP}{dp} = \binom{n}{y_{0}} (y_{0}p^{y_{0}−1}(1 − p)^{n−y_{0}} + (n − y_{0})p^{y_{0}}(1 − p)^{n−y_{0}−1}) .

Set this equal to zero and divide through by \binom{n}{y_{0}}p^{y_{0}−1}(1 − p)^{n−y_{0}−1} yields the equation:

0 = y_{0}(1 − p) + (n − y_{0})p

Solving for p we find p = \frac{y_{0}}{n}. This essentially has to be a local (and global) maximum since it is the only critical point. Note that the probability is always non-negative, but it equals zero when p = 0, 1. Hence on the interval 0 ≤ p ≤ 1 the critical point has to be a local maximum.

3.65. (a) We want to think of Y as binomial random variable, so using 3.33 (a),

E (\frac{Y}{n} ) = \frac{1}{n}E(Y) = \frac{1}{n} · np = p.

(b) Similarly, using the result of 3.33 (b)

V (\frac{Y}{n} ) = \frac{1}{n^{2}} V (Y) = \frac{npq}{n^{2}} = \frac{pq}{n} .

From this we see that as n → ∞, the variance of \frac{Y}{n} goes to zero.