Mathematics 375 – Probability and Statistics I Solutions for Problem Set 4 October 3, 2005

3.3/ The possible orders in which the 4 components are tested and the two defectives found are OODD, ODDO, DDOO, ODOD, DODO, DOOD (D - defective, O - OK). Assume each is equally likely. We have for Y = number of the test on which the second defective is found P (Y = 1) = 0, P (Y = 2) = 1/6 (DDOO), P (Y = 3) = 1/3 (ODDO, DODO) and P (Y = 4) = 1/2.

3.4/ Y = 0 means valve 1 fails, and either valve 2 or valve 3 fails. So by independence, P (Y = 0) = (.2)(.2 + .2 − .04) = .072; similarly, Y = 1 means valve 1 does not fail but one of 2, 3 does, or else valve 1 fails and both 2,3 do not. We have P (Y = 1) = (.8)(.2 + .2 − .04) + (.64)(.2) = .416. Then P (Y = 2) = 1 − (.072 + .416) = .512.

Alternate method – you can list out the 8 different possible outcomes for the 3 valves: each either works (W) or not (N). Then figure out how many open paths there are from A to B in each case. Add the appropriate probabilities to get P (Y = y) for y = 0, 1, 2. For instance, the cases where you get Y = 1 are WNN, WNW, WWN, NWW. The probabilities of each of these are P (W NN) = (.8)(.2)^{2} = .032, P (W NW) = P (W W N) = P (NW W) = (.8)^{2}(.2) = .128, so P (Y = 1) = .032 + 3(.128) = .416.

3.7/ a) P (Y = y) = \binom{3}{y}(.05)^{y}(.95)^{3−y} (this is a binomial random variable) c) P (Y > 1) = \binom{3}{2}(.05)^{2}(.95) + \binom{3}{3}(.05)^{3} = .00725.

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}

3.13/ Let C be the amount of the premium. We want the expected profit: (C − 15)(.98) + (C − 15 − 1000)(.02) = 50, so C = 85.

3.18/ If Y is the number of bottles with flaws out of the 2 P (Y = 0) = .9^{2} = .81, P (Y = 1) = 2(.9)(.1) = .18 and P (Y = 2) = .1^{2} = .01. So E(Y) = (.81)(0) + (.18)(1) + (.01)(2) = .2, E(Y^{2}) = (.81)(0)^{2}+(.18)(1)^{2}+(.01)(2)^{2} = .22 so V (Y) = E(Y^{2})−(E(Y))^{2} = .22 − .04 = .18. (Note: This is a binomial random variable, and we have derived general formulas for the expected value and variance. We’ll use them in the problems from the next section!)

3.22/ Let T be the total cost of testing and repairing. Then E(T) = 4 · 1/6 + 6 · 2/6 + 8 · 3/6 + 8 = 44/3.

3.26/ a) Since each judge will essentially select one of the three cups randomly, there is a p = 1/3 probability that they will pick the new formula B, and q = 2/3 probability that they will pick the old formula. Hence if Y is the number of judges who pick B, Y is binomial with probability mass function: P (Y = y) = \binom{4}{y}(1/3)^{y}(2/3)^{4−y}. b) 8/81 + 1/81 = 1/9. c) E(Y) = np = 4 · (1/3) = 4/3. d) V (Y) = npq = 4 · (1/3) · (2/3) = 8/9.

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.

3.29/ From the table with n = 15, p = .2, P (Y ≥ 10) ≐ 0 (i.e. very small). Using Maple, we can get a more precise value for ∑_{y=10}^{15} \binom{15}{y}(.2)^{y}(.8)^{15−y} ≐ .00011.

3.32/ a) (.8)^{5} = .32768. b) \binom{5}{4}(.6)^{4}(.4) ≐ .2592. c) (.7)^{5} + \binom{5}{1}(.7)^{4}(.3) ≐ .52822.

3.44/ Using the binomial probability table in the book with n = 20, p = .8: a) P (Y = 14) = P (Y ≤ 14) − P (Y ≤ 13) = .196 − .087 = .109. b) P (Y ≥ 10) = 1 − P (Y ≤ 9) ≐ .999 c) P (Y ≤ 16) = .589. d) E(Y) = np = 20(.8) = 16, V (Y) = npq = (20)(.8)(.2) = 3.2.

3.50/ a) By the geometric series sum formula,

\begin{aligned}∑_{y=1}^{∞} q^{y−1}p & = p + pq + pq^{2} + … \\ & = p · \frac{1}{1 − q} = 1. \end{aligned}

b) \frac{q^{y−1}p}{q^{y−2}p} = q < 1. This means P (Y = 1) is greater than P (Y = y) for all y > 1.

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}.

(Note Y > a+b implies Y > a, so the event (Y > a+b)∩(Y > a) is the same as Y > a+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”.