Mathematics 375 – Probability and Statistics I Solutions for Problem Set 4 October 3, 2003

3.10. Using the information given in the table,

\begin{aligned}E(Y) & = (1)(.4) + (2)(.3) + (3)(.2) + (4)(.1) = 2 \\ E(1/Y) & = (1)(.4) + (.5)(.3) + (.\overline{3})(.2) + (.25)(.1) ≐ .6417 \\ E(Y^{2} − 1) & = (1 − 1)(.4) + (4 − 1)(.3) + (9 − 1)(.2) + (16 − 1)(.1) = 4 \\ V (Y) & = E(Y^{2}) − (E(Y))^{2} = 5 − 4 = 1 \end{aligned}

3.17. There are 8 jacks or queens, 8 kings or aces, and 52 − 16 = 36 other cards in a standard 52-card deck. The card is drawn at random so the probability of getting a jack or a queen is \frac{8}{52}, and so forth. Hence the expected value of the winnings in one game is

E(W) = (15)\frac{8}{52} + (5)\frac{8}{52} + (−4)\frac{36}{52} = \frac{4}{13} ≐ .31

(In other words, on average you “expect” to win 31 cents on each game. “Sweet!!!”)

3.19. Consider the random variable Y = number of contracts assigned to firm I. Since the contracts are assigned randomly to the 3 firms, the probability distribution function (pdf) for Y is:

p(0) = P (Y = 0) = \frac{2}{3} · \frac{2}{3} = \frac{4}{9}

p(1) = P (Y = 1) = \frac{2}{3} · \frac{1}{3} + \frac{1}{3} · \frac{2}{3} = \frac{4}{9}

p(2) = P (Y = 2) = \frac{1}{3} · \frac{1}{3} = \frac{1}{9}

(Comment: Note that in the language from the next section, Y is a binomial random variable with success probability p = 1/3.) The profit is the random variable 90, 000Y, so

E(90, 000Y) = (90, 000)\frac{4}{9} + (180, 000)\frac{1}{9} = 60, 000.

The expected profit for each of the three firms is the same, so if one person owns I and II, the expected profit for that person is 60, 000 + 60, 000 = 120, 000.

3.21. Let L be the random variable representing the loss expected from a single policy. The given information determines the pdf for L (note it says, ignore all but total loss and half loss). If the value of the policy is 85,000, then

E(L) = 85, 000P (total loss) + 42, 500P (half loss) = (85, 000)(.001) + (42, 500)(.01) = 510.

The insurance company should charge $510 to come out even (on average).

3.27. Let Y be the number of components that fail in less than 1000 hours. This is a binomial random variable with p = .2 and q = 1 − p = .8. For a), P (Y = 2) = \binom{4}{2}(.2)^{2}(.8)^{2} ≐ .1536. For b), since the problem says the system can operate if any two of the components are operating, the probability that the system is operating after 1000 hours is P (Y ≤ 2) = ∑_{k=0}^{2} \binom{4}{k}(.2)^{k}(.8)^{4−k} ≐ .9728.

3.28. The number Y that recover is a binomial random variable with p = .8 and q = .2.

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

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

c) P (14 ≤ Y ≤ 18) = ∑_{k=10}^{20} \binom{20}{k}(.8)^{k}(.2)^{20−k} ≐ .844.

d) P (Y ≤ 16) = ∑_{k=0}^{16} \binom{20}{k}(.8)^{k}(.2)^{20−k} ≐ .589 Note: these can be computed by calculator, using Maple, or using the Table 1 in Appendix III of the text. The table gives P (Y ≤ k) for each k, so to use the table for c) for instance you would take P (Y ≤ 18) − P (Y ≤ 13) = .931 − .087.

3.31/ The number of families who qualify is a binomial random variable Y with p = .7. For a), the probability P (Y = 5) = \binom{5}{5}(.7)^{5} ≐ .1681. For b), P (Y ≥ 4) = \binom{5}{4}(.7)^{4}(.3) + \binom{5}{5}(.7)^{5} ≐ .5282.

3.36/ The number of radar sets Y that detect the missile is a binomial random variable with p = .9. The probability that exactly 4 sets detect the missile is P (Y = 4) = \binom{5}{4}(.9)^{4}(.1) ≐ .32805. The probability that at least one does is P (Y ≥ 1) = 1 − P (Y = 0) = 1 − (.1)^{5} = .99999. For b), the general pattern is that with n radar sets, P (Y ≥ 1) = 1 − (.1)^{n}. This is ≥ .999 for all n ≥ 3. So (only) 3 radar sets are necessary. (Of course, you don’t want to be in the .001 case where the missile goes undetected!!!)

3.39/ We follow the pattern from the proof of the formula for the expected value of Y (Y −1):

E(Y (Y − 1)(Y − 2)) = ∑_{y=0}^{n} y(y − 1)(y − 2)\binom{n}{y}p^{y}q^{n−y}

= ∑_{y=0}^{n} y(y − 1)(y − 2)\frac{n!}{y(y − 1)(y − 2)(y − 3)!(n − y)!}p^{y}q^{n−y}

= n(n − 1)(n − 2)p^{3} ∑_{y=3}^{n} \frac{(n − 3)!}{(y − 3)!((n − 3) − (y − 3))!}p^{y−3}q^{(n−3)−(y−3)}

= n(n − 1)(n − 2)p^{3} ∑_{z=0}^{n−3} \frac{(n − 3)!}{z!((n − 3) − z)!}p^{z}q^{(n−3)−z}

= n(n − 1)(n − 2)p^{3} (where we wrote z = y−3 to reindex the sum, and then used the fact that the sum was sum of the pdf-values for a binomial random variable with n − 3 trials and success probability p. By the linearity of the expectation:

n(n − 1)(n − 2)p^{3} = E(Y^{3} − 3Y^{2} + 2Y) = E(Y^{3}) − 3E(Y^{2}) + 2E(Y)

Since we already know E(Y^{2}) = n(n − 1)p^{2} + np (from derivation of formula for V (Y) for a binomial random variable) and E(Y) = np, we get

E(Y^{3}) = n(n − 1)(n − 2)p^{3} + 3n(n − 1)p^{2} + np.

3.51/ Let Y be the number of the interview on which the first one with advanced training is found. This is a geometric random variable with p = .3. Hence P (Y = 5) = (.7)^{4}(.3) ≐ .07203.

3.52/ Continuing from 3.51, we have E(Y) = 1/.3 = 10/3 = 3.\overline{3} by the formula we proved in class for the expected value of a geometric random variable.

3.55/ For a geometric random variable Y,

P (Y > a) = ∑_{y=a+1}^{∞} q^{y−1}p = \frac{pq^{a}}{1 − q} = q^{a}

(where we used the geometric series sum formula – this is a geometric series with first term q^{a}p and ratio q). Similarly, for part b) using the definition of conditional probability and part a),

P (Y > a + b|Y > a) = P (Y > a + b)/P (Y > a) = q^{a+b}/q^{a} = q^{b}.

Comment: This is the same as P (Y > b). In other words, for geometric random variables, the probability that Y > a + b given that Y > a (i.e. the probability that the first success occurs after trial a + b, given that the first success occurs after trial a) is the same as the probability that the first success occurs after trial b. Geometric random variables thus are “memoryless” in the sense that starting from the a-th trial (with no previous successes), the probability that the first success occurs more than b trials later is the same as if we started from the beginning with the 0th trial. The previous non-successes don’t affect the probability of later successes – no “memory”.

3.57/ Y is the number on which the first audit with substantial errors is found – geometric with p = .9. For a), P (Y = 3) = (.1)^{2}(.9) = .009. For b), P (Y ≥ 3) = P (Y > 2) = (.1)^{2} = .01 by problem 55.

3.58/ Use the formulas for geometric random variables from the text Theorem 3.8 or class notes: E(Y) = 1/.9 = 10/9. σ = \sqrt{(1 − .9)/(.9)^{2}} ≐ .35.