Mathematics 375–Probability and Statistics I Solutions for Problem Set 3 September 26, 2003

A) 2) For large numbers of runs (say ≥ 10000), the fraction \frac{wins}{totalgames} should be roughly .493. 3) a) The probability of winning immediately is the probability of rolling a total of 7 or 11, which is \frac{6}{36} + \frac{2}{36} = \frac{2}{9}. The probability of losing immediately is the probability of rolling a 2, 3, or 12, which is \frac{1}{36} + \frac{2}{36} + \frac{1}{36} = \frac{1}{9}. b) Let A be the event that you win, F 4 be the event that you roll a 4 on the first roll. The conditional probability P (A|F 4) can be analyzed like this: A ∩ F 4 is the union of “subevents” A_{k}:

A_{2} = 4 is rolled on 2nd roll,
A_{3} = 4 is rolled on 3rd roll and second roll was not a 7,
A_{4} = 4 is rolled on 4th roll and second and third rolls are not 7’s,
⋮

These are mutually exclusive (no overlaps), but any number of rolls can happen before the next 4 comes up. Hence

P (A ∩ F 4) = ∑_{k=2}^{∞} P (A_{k}) = ∑_{k=2}^{∞} (\frac{3}{36}) (\frac{27}{36})^{k−2}

(the \frac{3}{36} is the probability of rolling a 4; the \frac{27}{36} is the probability of rolling anything other than a 4 or a 7).

Then reasoning similarly, let F 5, F 6, F 8, F 9, F 10 be the events corresponding to the different first rolls in this case.

\begin{aligned}P (A ∩ F 10) & = P (A ∩ F 4) \\ P (A ∩ F 5) & = P (A ∩ F 9) = ∑_{k=2}^{∞} P (A_{k}) = ∑_{k=2}^{∞} (\frac{4}{36}) (\frac{26}{36})^{k−2} \\ P (A ∩ F 6) & = P (A ∩ F 8) = ∑_{k=2}^{∞} P (A_{k}) = ∑_{k=2}^{∞} (\frac{5}{36}) (\frac{25}{36})^{k−2} \end{aligned}

Then by the Law of Total Probability

P (A) = P (win immediately) + P (A|F 4)P (F 4) + P (A|F 5)P (F 5) + P (A|F 6)P (F 6)

+ P (A|F 8)P (F 8) + P (A|F 9)P (F 9) + P (A|F 10)P (F 10)

= \frac{2}{9} + 2 ∑_{k=2}^{∞} (\frac{3}{36})^{2} (\frac{27}{36})^{k−2} + 2 ∑_{k=2}^{∞} (\frac{4}{36})^{2} (\frac{26}{36})^{k−2}

+ 2 ∑_{k=2}^{∞} (\frac{5}{36})^{2} (\frac{25}{36})^{k−2}

= \frac{244}{495} ≐ .49283

(The sums are geometric series and can be summed by the usual formula:

∑_{ℓ=0}^{∞} cr^{ℓ} = \frac{c}{1 − r} .)

The extra credit part is entirely similar, but you must use the new probabilities for each dice roll. The formula from the Law of Total Probability is still valid, but now P (F 7) = \frac{4}{64} + \frac{2}{16} = \frac{3}{16}, P (F 11) = \frac{1}{32} P (F 4) = P (F 10) = \frac{2}{32} + \frac{1}{64} = \frac{5}{64}, and so on. The result for the winning probability is \frac{7581}{14960} ≐ .507. In other words, “Buggsy is going to be very happy”(!)

2.61/ P (A) = .4 is the sum of the entries on the first row (but not the total, of course!), P (B) = .37 is the sum of the entries in the first column, P (A ∩ B) = .1 is the entry in the upper left, P (A ∪ B) = .67 is the sum of the three entries excluding lower right (and excluding totals), P (\overline{A}) = 1 − P (A) = .6, P (\overline{A ∪ B}) = 1 − P (A ∪ B) = .67, P (\overline{A ∩ B}) = 1 − P (A ∩ B) = .9, P (A|B) = P (A ∩ B)/P (B) = .1/.37 ≐ .2703, and P (B|A) = .1/.4 = .25.

2.63/ The events are mutally exclusive, so P (A ∩ B) = 0. Hence if P (B) > 0, then 0 = P (A ∩ B) ≠ P (A)P (B). So the events are not necessarily independent.

2.64/ Since A ⊂ B, we have P (A ∩ B) = P (A). Hence if P (B) < 1, it cannot be true that P (A ∩ B) = P (A)P (B). So the events are not necessarily independent.

2.66/ The general rule for triple unions is derived from the sum rule in the text:

\begin{aligned}P (A_{1} ∪ A_{2} ∪ A_{3}) & = P (A_{1}) + P (A_{2}) + P (A_{3}) − P (A_{1} ∩ A_{2}) − P (A_{1} ∩ A_{3}) − P (A_{2} ∩ A_{3}) \\ & + P (A_{1} ∩ A_{2} ∩ A_{3}) \end{aligned}

We are given that P (A_{2} ∩ A_{3}) = 0, and that implies P (A_{1} ∩ A_{2} ∩ A_{3}) = 0 since A_{1} ∩ A_{2} ∩ A_{3} ⊂ A_{2} ∩ A_{3}. So the last two terms above are both zero. Furthermore, we are given P (A_{1} ∩ A_{3}) = P (A_{1} ∩ A_{2}), so

P (A_{1} ∪ A_{2} ∪ A_{3}) = P (A_{1}) + P (A_{2}) + P (A_{3}) − 2P (A_{1} ∩ A_{2}),

which is what we had to show.

2.68/ Let F_{i} be the event that the ith person is telling the truth, but the lie detector indicates that he/she is lying. Part a) asks for P (F_{1} ∩ F_{2} ∩ F_{3}) = (.05)^{3} = .000125, since the F_{i} are independent events. Part b) asks for P (F_{1} ∪ F_{2} ∪ F_{3}), which we can compute using the formula from 2.66:

P (F_{1} ∪ F_{2} ∪ F_{3}) = 3(.05) − 3(.0025) + .000125 = .142625.

(Alternate Method: We could also derive the answer for b) by using the complement rule:

P (F_{1} ∪ F_{2} ∪ F_{3}) = 1 − P (\overline{F_{1}} ∩ \overline{F_{2}} ∩ \overline{F_{3}}) = 1 − (.95)^{3} = .142625.

(Note: \overline{F_{i}} is the event: “The person is telling the truth and the lie detector agrees.”)

2.69/ Let H represent a hit and M a miss on one attempt, and let L, R represent which hand is used. The given information says that P (H|R) = .7 (so P (M|R) = 1 − .7 = .3) and P (H|L) = .4 (so P (M|L) = .6). The winning sequences of outcomes in the three attempts are HHM, MHH, and HHH. Starting on the right hand and using the independence, P (HHM) = P (H|R)P (H|L)P (M|R) = (.7)(.4)(.3), P (MHH) = P (M|R)P (H|L)P (H|R) = (.3)(.4)(.7), and P (HHH) = P (H|R)P (H|L)P (H|R), which = (.7)(.4)(.7). The sum gives the total probability of winning: .364.

2.73/ Let C_{i} be the event: relay i activates correctly. We have P (C_{i}) = .9, and the C_{i} are independent. For a), the probability that at least one open path from A to B exists is P (C_{1} ∪ C_{2} ∪ C_{3}) = 3 × .9 − 3 × .81 + .729 = .999 using the formula from 2.66 above. For b) we want P (C_{1}|C_{1} ∪ C_{2} ∪ C_{3}) = P (C_{1})/P (C_{1} ∪ C_{2} ∪ C_{3}), using the definition of the conditional probability. This gives .9/.999 ≐ .9009.

2.80/ Using the DeMorgan and sum laws we have

1 − P (A ∩ B) = P (\overline{A ∩ B})
= P (\overline{A} ∪ \overline{B})
= P (\overline{A}) + P (\overline{B}) − P (\overline{A} ∩ \overline{B})
= P (\overline{A}) + P (\overline{B}) − P (\overline{A ∪ B}).

So rearranging,

P (A ∩ B) = 1 − P (\overline{A}) − P (\overline{B}) + P (\overline{A ∪ B}) ≥ 1 − P (\overline{A}) − P (\overline{B}),

(since P (\overline{A ∪ B}) ≥ 0), which is what we had to show.

2.84/ Using 2.80 twice:

P (A ∩ B ∩ C) = P ((A ∩ B) ∩ C)
≥ 1 − P (\overline{A ∩ B}) − P (\overline{C})
= P (A ∩ B) − P (\overline{C})
≥ 1 − P (\overline{A}) − P (\overline{B}) − P (\overline{C})

which is what we had to show. Comment: The same argument extends to any finite number of sets:

P (∩_{i=1}^{n}A_{i}) ≥ 1 − ∑_{i=1}^{n} P (\overline{A_{i}}).

2.87/ Consider following events: M = person sees magazine ad, T = person sees TV ad, S = M ∪ T person sees ad in at least one place, and B = person buys. The given information is first that P (M) = .02, P (T) = .2, P (M ∩ T) = .01, so by the sum rule we also know P (S) = P (M ∪ T) = .02 + .2 − .01 = .21. We are also given P (B|S) = .\overline{3}, and P (B|\overline{S}) = .1. By the Law of Total Probability,

P (B) = P (B|S)P (S) + P (B|\overline{S})P (\overline{S}) = (.\overline{3})(.21) + (.1)(1 − .21) = .149

2.93/ Let P be the event that a car passes inspection, and let T_{i} be the event that team i is chosen to do the inspection. The given information says P (P |T_{1}) = 1 and P (P |T_{2}) = 0. The sample space for the four cars is the set of ordered 4-tuples in {T_{1}, T_{2}}^{4} (indicating which team is selected for each car). a) The probability that three will be rejected is P (E) where E = {(T_{1}, T_{2}, T_{2}, T_{2}), (T_{2}, T_{1}, T_{2}, T_{2}), (T_{2}, T_{2}, T_{1}, T_{2}), (T_{2}, T_{2}, T_{2}, T_{1})}. So P (E) = 4/16 = 1/4. For b), we have P ({T_{1}, T_{1}, T_{1}, T_{1}}) = 1/16.

2.99/ Let D be the event: person has the disease, and let T be the event: test is positive (indicates disease is present). We are given P (T |D) = .9 and P (\overline{T} |\overline{D}) = .9 (so P (T |\overline{D}) = .1 – the probability that the test will be positive even though the patient does not have the disease – or “false positive” test results). We also know P (D) = .01. For a random person, we can find P (D|T) using Bayes’ Rule:

P (D|T) = \frac{P (T |D)P (D)}{P (T |D)P (D) + P (T |\overline{D})P (\overline{D})} = \frac{(.9)(.01)}{(.9)(.01) + (.1)(.99)} = .083.

Comment: This should be an absolutely shocking result. It says a “90% reliable” test in this sense is virtually worthless. The probability that the disease is present, given that the test was positive is only .083(!) The “moral” is that tests have to be much more reliable than this to give useful information(!)

2.102/a) Since P (A|B) = P (A|\overline{B}), by the Law of Total Probability,

P (A) = P (A|B)P (B) + P (A|\overline{B})P (\overline{B})
= P (A|B)(P (B) + P (\overline{B}))
= P (A|B)

Then A and B are independent according to Definition 2.10. For b), given P (A|C) > P (B|C) and P (A|\overline{C}) > P (B|\overline{C}), by the Law of Total Probability,

P (A) = P (A|C)P (C) + P (A|\overline{C})P (\overline{C})
> P (B|C)P (C) + P (B|\overline{C})P (\overline{C})
= P (B).

2.107/We have A ∪ B = (AΔB) ∪ (A ∩ B), and (AΔB) ∩ (A ∩ B) = ∅. Hence by the Sum Rule

P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = P (AΔB) + P (A ∩ B)

Rearranging yields:

P (AΔB) = P (A) + P (B) − 2P (A ∩ B).

3.1/ Given P (A) = .4, P (B) = .5, and P (\overline{A ∪ B}) = .2. Then by the Sum Rule P (A ∪ B) = .5+.4−P (A∩B) = 1−.2 = .8 Hence P (A∩B) = .1. Therefore the probability distribution for the random variable Y = number of impurities found is

\begin{aligned}p(0) & = P (Y = 0) = P (\overline{A ∪ B}) = .2 \\ p(1) & = P (Y = 1) = P (AΔB) = .7 \\ p(2) & = P (Y = 2) = P (A ∩ B) = .1 \end{aligned}

(all other values 0). The value for p(1) is computed using the formula from 2.107, or by taking 1 − (.2 + .1).

3.3/ The random variable Y is the number of the test on which the second defective is found. We can find the probability distribution by considering the different possible sequences in which the 2 OK and 2 defective components could be found:

\begin{aligned}DDOO & ⇒ Y = 2 \\ DODO & ⇒ Y = 3 \\ DOOD & ⇒ Y = 4 \\ ODDO & ⇒ Y = 3 \\ ODOD & ⇒ Y = 4 \\ OODD & ⇒ Y = 4 \end{aligned}

Since the components are selected randomly any sequence has probability 1/6, and the distribution is:

\begin{aligned}p(2) & = P (Y = 2) = 1/6 \\ p(3) & = P (Y = 3) = 2/6 = 1/3 \\ p(4) & = P (Y = 4) = 3/6 = 1/2 \end{aligned}

(all other values 0).