Mathematics 375 – Probability Theory Solutions for Problem Set 3 September 23, 2011

2.85. Since S = B∪\overline{B} and B∩\overline{B} = ∅, we have A = (A∩B)∪(A∩\overline{B}) and P (A) = P (A∩B)+P (A∩\overline{B}). Since A and B are independent, P (A ∩ B) = P (A)P (B). Substituting for P (A), we obtain:

P (A ∩ B) = (P (A ∩ B) + P (A ∩ \overline{B}))P (B),

so after algebraic rearrangement

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

Since P (A ∩ B) = P (A)P (B) again, this simplifies to

P (A)P (\overline{B}) = P (A ∩ \overline{B}),

which shows that A and \overline{B} are independent. The answer is yes for the second question. From P (\overline{A} ∩ \overline{B}) + P (A ∩ \overline{B}) = P (\overline{B}),

P (\overline{A} ∩ \overline{B}) = P (\overline{B}) − P (A ∩ \overline{B})
= P (\overline{B}) − P (A)P (\overline{B}) by the first part
= P (\overline{B})(1 − P (A))
= P (\overline{B})P (\overline{A})

This shows \overline{A} and \overline{B} are independent.

2.86. (a) No it is not possible. That would say P (A ∪ B) = .8 + .7 − .1 = 1.4. But P (A ∪ B) ≤ 1.

(b) The smallest value is P (A ∩ B) = .5. (c) No it is not possible, P (B) = .7 and A ∩ B ⊆ B, so P (A ∩ B) ≤ P (B). (d) The largest value is P (A ∩ B) = .7.

2.90. (a) Let A be the event that the sky diver is injured on the first jump, and B the event that she is injured on the second jump. Under the independence assumption

P (A ∩ B) = P (A)P (B) = \frac{1}{50} · \frac{1}{50} = \frac{1}{2500}.

Hence the probability that she is injured (on at least one jump) is:

P (A ∪ B) = \frac{1}{50} + \frac{1}{50} − \frac{1}{2500} = \frac{99}{2500}.

(b) No, your friend has it wrong. Note that the probability of injury after two jumps is already slightly less than \frac{2}{50}. After 50 jumps, the probability that you have been injured in at least one jump can be computed like this. The event we want is the complement of the event indicating that all jumps occurred without injury. So the probability is

1 − (\frac{49}{50})^{50} ≐ .636

(that is you have about a 64% chance of being injured at least once).

2.91. If events A and B are mutually exclusive, then P (A ∩ B) = 0. If P (A) = .4 and P (B) = .7, then P (A ∩ B) ≥ .1 (as in 2.86 above). So A and B are not mutually exclusive in this case. If P (A) = .4 and P (B) = .3, then P (A ∩ B) = 0 is possible (but not always the case, of course!).

2.92. (a) Let A_{1}, A_{2}, A_{3} be the events that the three employees are telling the truth but the lie detector test indicates that each is lying. Because of independence, the probability that all three events occur is

P (A_{1} ∩ A_{2} ∩ A_{3}) = (.05)^{3} = .000125

(b) The probability that at least one occurs is the complementary probability for the event that the lie detector never gives an incorrect result. This is 1 − (.95)^{3} ≐ .143. This can also be computed by the additive rule for unions of more than two events: By the formula from page 58 in the text,

P (A_{1} ∪ A_{2} ∪ A_{3}) = 3(.05) − 3(.05)^{2} + (.05)^{3} ≐ .143.

Either way, there is about a 14% chance that the lie detector test is wrong at least once.

2.94. (a) This is P (A ∪ B) = .95 + .90 − .88 = .97. (The combination is more reliable than either separately.) (b) The probability that the smoke is undetected is P (\overline{A ∪ B}) = 1 − .97 = .03.

2.95. (a) From the additive law (Theorem 2.6), .4 = .2 + .3 − P (A ∩ B), so P (A ∩ B) = .1. (b) By DeMorgan’s Laws, P (\overline{A} ∪ \overline{B}) = P (\overline{A ∩ B}) = 1 − .1 = .9.

(c) Similarly, P (\overline{A} ∩ \overline{B}) = P (\overline{A ∪ B}) = 1 − .4 = .6.

(d) P (\overline{A}|B) = \frac{P (\overline{A}∩B)}{P (B)}. The numerator can be computed since we know P (A ∩ B) = .1 and P (A ∩ B) + P (\overline{A} ∩ B) = P (B) = .3. So the probability we want is \frac{.2}{.3} = \frac{2}{3}.

2.97. (a) The probability of current flowing is

P (1 ∪ 2 ∪ 3) = P (1) + P (2) + P (3) − P (1 ∩ 2) − P (1 ∩ 3) − P (2 ∩ 3) + P (1 ∩ 2 ∩ 3)
= .9 + .9 + .9 − .81 − .81 − .81 + .729
= .999

(Alternate method: The only way current can fail to flow is if all three relays fail, which happens with probability (.1)^{3} = .001. So the complementary probability is .999.)

(b) If relay 1 functioned properly, then current did flow, so P (flow ∩ 1) = P (1) = .9. Hence P (1|flow) = \frac{.9}{.999} = .\overline{900}.

2.99. Since A and B are independent, P (A ∩ B) = P (A)P (B). So

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

Solving for P (A), we get

P (A) = \frac{1 − a − b}{1 − b},

as desired.

2.114. The given information says that one person is guilty and will lie when asked “did you do it?” and the other is innocent and tells the truth when asked that same question. In the products in the following, in each case the first factor corresponds to the innocent person, and the second one corresponds to the guilty person. (a) The probability that the lie detector gives a positive reading for both (i.e. indicates both suspects are lying) is (because of independence)

(.10)(.95) = .095.

(b) Similarly, (.90)(.95) = .855. (c) (.10)(.05) = .005. (d) Using the Additive Law, .10 + .95 − (.10)(.95) = .955.

2.115. Because of independence of the games, the probability is (.75)^{4} ≐ .32.

2.120. To me, the most straightforward way to attack these is to list out the different sequences in which the 2 defective fridges (D’s) and the 4 good ones (G’s) could be found in the testing. I’m assuming the fridges are not distinguishable here:

DDGGGG,DGDGGG, DGGDGG, DGGGDG, DGGGGD,
GDDGGG,GDGDGG, GDGGDG, GDGGGD,
GGDDGG,GGDGDG, GGDGGD,
GGGDDG,GGGDGD,
GGGGDD.

These \binom{6}{2} = 15 possible sequences would be equally likely, given that the fridges are chosen at random for the testing. Of the 15, exactly 3 have the second D in the 4th position –

{DGGDGG, GDGDGG, GGDDGG}.

So this probability is \frac{3}{15} = \frac{1}{5}. (b) The second D is found no later than the 4th test in 6 of the cases

– the three from part (a) and

{DDGGGG, DGDGGG, GDDGGG}

So this probability is \frac{6}{15} = \frac{2}{5}. (c) The event A = exactly one defective found in first two tests consists of

A = {DGDGGG, DGGDGG, DGGGDG, DGGGGD, GDGDGG, GDDGGG, GDGGDG, GDGGGD}

Then B = remaining defective found in 3rd or 4th test gives

B ∩ A = {DGDGGG, DGGDGG, GDDGGG, GDGDGG}.

So

P (B|A) = \frac{P (B ∩ A)}{P (A)} = \frac{\frac{4}{15}}{\frac{8}{15}} = \frac{1}{2}.

2.125. Let A be the event that the test indicates presence of the disease, and B be the event that the disease is actually present. With a randomly chosen person, P (B) = .01 and P (\overline{B}) = .99. The given information says P (A | B) = .9 and P (\overline{A} | \overline{B}) = .9, so P (A | \overline{B}) = .1. But we are interested in the conditional probability P (B | A) – the chance that the person has the disease, given that the test was positive. Using the partition B ∪ \overline{B} = U, Bayes’ rule says

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

= \frac{(.9)(.01)}{(.9)(.01) + (.1)(.99)}

≐ .083.

Your reaction should be: “What??? That can’t possibly be right.” This should seem absurdly low, and completely contradictory to what the “90% reliability” given information would seem to be saying. But it is correct and a standard example of how the results of diagnostic testing can give misleading results. The problem is the large size of the second term on the bottom relative to the first term. The term P (A | \overline{B})P (\overline{B}) = P (A ∩ \overline{B}) represents the chance of a false positive result – a case where the person does not have the disease, but the test comes back positive anyway. The large number of false positives swamps the true positive results that the test is designed to produce. Comment: The pattern seen in this question has major implications concerning quality and affordability of health care!

2.128. (a) By LTP, we always have

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

If P (A | \overline{B}) = P (A | B), then we substitute into the second term in the right and collect terms like this:

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

This is one of the three equivalent forms of the statement “A and B are independent.” (See Definition 2.10.) (b) By the LTP,

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

If P (A | C) > P (B | C) and P (A | \overline{C}) > P (B | \overline{C}), then since P (C), P (\overline{C}) ≥ 0, we have

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

2.129. Let A be the event that the randomly chosen person is a woman, and B be the event that the person is a man. Let C be the event that he or she reacts positively to the mysterious “circumstances” in the question. We have P (A) = \frac{15}{20} = .75 and P (B) = .25. Moreover, P (C | A) = .7 (so P (\overline{C} | A) = .3) and P (C | B) = .4, (so P (\overline{C} | B) = .6). We want P (B | \overline{C}). By Bayes’ rule:

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

= \frac{(.6)(.25)}{(.6)(.25) + (.3)(.75)}

= .4.

2.134. Again, we use Bayes’ rule. Let A, B be the events that the corresponding teaching methods were used and F be the event that the person fails the test. We have P (F | A) = .2 and P (F | B) = .3. Moreover, P (A) = .7 and P (B) = .3, so A and B form the partition of the sample space in this case.

P (A | F) = \frac{P (F | A)P (A)}{P (F | A)P (A) + P (F | B)P (B)}

= \frac{(.2)(.7)}{(.2)(.7) + (.1)(.3)}

≐ .824.

3.1. Let Y = number of impurities found in one well. Since 20% of the wells had neither, 80% had at least one. Thus, by the additive law, the percent that had both was (40 + 50 − 80)% = 10%. Then P (Y = 0) = .2, P (Y = 1) = .7 and P (Y = 2) = .1.

3.4. There are two open paths only when all three valves open correctly. This happens with probability (.8)^{3} = .512. There are no open paths when valve 1 fails, and valve 2 or valve 3 or both fail. This happens with probability (.2)(.2 + .2 − (.2)^{2}) = .072. Since only 0,1, or 2 paths can occur, the probability of exactly one path open is then 1 − (.512 + .072) = .416. The probability distribution for Y = number of open paths is P (Y = 0) = .072, P (Y = 1) = .416, P (Y = 2) = .512.

3.5. Say the correct assignment is ABC in that order. The possible assignments the child makes are then ABC, ACB, BAC, BCA, CAB, CBA. The first gives all three correct, the next three give one correct, and the last two give no correct. Hence if Y = the exact number correct if one of the 6 assignments is chosen at random, we have P (Y = 0) = \frac{1}{3}, P (Y = 1) = \frac{1}{2}, P (Y = 2) = 0, and P (Y = 3) = \frac{1}{6}. (Note that it is not possible to get exactly 2 correct, since then the third would have to be correct too.)

3.7. The answer in the book corresponds to an interpretation where you think of the balls as being distinguishable (perhaps differently colored, or different sizes). This means that it matters which balls go into which bowls, not just the total numbers that end up in each bowl. One way to represent the sample space is by giving a collection of triples giving the bowl into which the corresponding ball is placed. For instance (1, 1, 2) would indicate balls 1 and 2 go into bowl 1 and ball 3 goes into bowl 2. In this case, bowl 3 is empty. There are 3^{3} = 27 sample points obtained this way. There are 3 ways to have two empty bowls (sample points (1, 1, 1), (2, 2, 2), (3, 3, 3)), 6 ways to have no empty bowls (sample points (1, 2, 3) and all permutations), and hence 18 ways to have 1 empty bowl. If Y = number of empty bowls, this gives P (Y = 0) = \frac{6}{27} = \frac{2}{9}, P (Y = 1) = \frac{18}{27} = \frac{2}{3}, and P (Y = 2) = \frac{3}{27} = \frac{1}{9}.

Note: I also gave full credit for solutions that treated the balls as indistinguishable. If you thought about it that way, then there are only 10 different outcomes, corresponding to the ways of splitting the 3 balls over the 3 bowls. Here you could represent the sample space by ordered triples with one entry for each bowl, giving the number of balls that end up in that bowl:

S = {(3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (2, 0, 1), (1, 2, 0), (0, 2, 1), (1, 0, 2), (0, 1, 2), (1, 1, 1)}

Note that if Y = number of empty bowls again, then the probabilities end up being P (Y = 2) = 3/10, P (Y = 1) = 6/10, and P (Y = 0) = 1/10.

Comment: It is sometimes difficult to state problems in this subject in ways that are totally unambiguous!

3.8. The question asks you to compute the probabilities for the numbers of cells in the second generation after the start. We start from a single cell (generation 0, so to speak). If that dies then there are no cells in generations 1 or 2. If the single cell splits, but both cells in generation 1 die, then there is another way for generation 2 to have 0 cells: P (Y = 0) = .1 + (.9)(.1)^{2} = .109. If both cells in generation 1 split then we have 4 cells in generation 2: P (Y = 4) = (.9)^{3} = .729. Finally, if the original cell split to form 2 cells in generation 1, but only one of those split and the other one died, there are two cells in generation 2 and P (Y = 2) = (.9)((.1)(.9) + (.9)(.1)) = .162.

3.9. We are supposed to assume that the auditor finds all the incorrect entries that are present. So this is really just the same as computing the probabilities of the different numbers of errors that could be made in the three entries that are tested. If Y = number of incorrect entries out of the three, P (Y = 0) = (.95)^{3} = .857375, P (Y = 1) = 3(.05)(.95)^{2} = .135375, P (Y = 2) = 3(.05)^{2}(.95) = .007125 and P (Y = 3) = (.05)^{3} = .000125. (Comment: This is a binomial random variable. We will study these in detail later in Chapter 3.) (b) The probability histogram is similar to the relative frequency histograms constructed in Chapter 1. (c) P (Y > 1) = P (Y = 2) + P (Y = 3) = .0725.