Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 3 September 24, 2009

2.71. (a) P (A|B) = \frac{P (A∩B)}{P (B)} = \frac{1}{3}.

(b) P (B|A) = \frac{P (A∩B)}{P (A)} = \frac{1}{5}

(c) We have P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = .7. Therefore, P (A|A ∪ B) = \frac{P (A)}{P (A∪B)} = \frac{5}{7}.

(d) P (A|A ∩ B) = 1, since (A ∩ B) ∩ A = A ∩ B.

(e) P (A ∩ B|A ∪ B) = \frac{1}{7}.

2.72. (a) P (A ∩ M) = .24 while P (A) = .6 and P (M) = .4. Since P (A ∩ M) = P (A)P (M), the events A, M are independent (see Definition 2.10).

(b) P (\overline{A} ∩ F) = .24 while P (\overline{A}) = .4 and P (F) = .6. Since P (\overline{A} ∩ F) = P (\overline{A})P (F), these events are also independent.

2.74. (a) P (A ∩ D) = .2, while P (A) = .61 and P (D) = .3 since .2 ≠ (.61)(.3), A and D are not independent.

(b) P (B ∩ D) = .09 = P (B)P (D) = (.3)(.3). So B and D are independent.

(c) P (C ∩ D) = .01, but P (C)P (D) = (.09)(.3) = .027. So C and D are not independent.

2.75. (a) Let A be the event that the first two cards dealt are spades, so P (A) = \frac{13}{52} · \frac{12}{51} = \frac{1}{17}. Next, let B be the event that cards 3, 4, 5 are also spades. We want

P (B|A) = \frac{P (B ∩ A)}{P (A)} = \frac{\frac{13}{52} · \frac{12}{51} · \frac{11}{50} · \frac{10}{49} · \frac{9}{48}}{\frac{13}{52} · \frac{12}{51}} = \frac{33}{3920} ≐ .008.

(b) Similarly, let C be the event that the first three cards are spades, and D be the event that cards 4, 5 are spades.

P (D|C) = \frac{P (D ∩ C)}{P (C)} = \frac{\frac{13}{52} · \frac{12}{51} · \frac{11}{50} · \frac{10}{49} · \frac{9}{48}}{\frac{13}{52} · \frac{12}{51} · \frac{11}{50}} = \frac{15}{392} ≐ .04.

(c) Similarly, let F be the event that the first four cards are spades, and F be the envent that card 5 is a spade.

P (F |E) = \frac{P (F ∩ E)}{P (E)} = \frac{\frac{13}{52} · \frac{12}{51} · \frac{11}{50} · \frac{10}{49} · \frac{9}{48}}{\frac{13}{52} · \frac{12}{51} · \frac{11}{50} · \frac{10}{49}} = \frac{9}{48} ≐ .19.

2.81. We are given that P (A), P (B) > 0 and P (A) < P (A|B). By the definition of conditional probability, this last statement says P (A) < \frac{P (A∩B)}{P (B)}, so since P (B) > 0, P (A)P (B) < P (A ∩ B).

Now divide this statement through by P (A): P (B) < \frac{P (A∩B)}{P (A)}. But this says P (B) < P (B|A), which is what we wanted to show.

2.84. “At least one A_{i}” means A_{1} ∪ A_{2} ∪ A_{3}. The general formula for the probability of a 3-fold union is given in the text on page 58. In the terms on the right, we know P (A_{2} ∩ A_{3}) = 0, and it follows that P (A_{1} ∩ A_{2} ∩ A_{3}) = 0, since A_{1} ∩ A_{2} ∩ A_{3} ⊂ A_{2} ∩ A_{3}. Moreover, we are given that P (A_{1} ∩ A_{2}) = P (A_{1} ∩ A_{3}). The required formula follows.

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.

2.88. (a) It is possible for P (A ∩ B) = .1 if P (A ∪ B) = .8.

(b) The smallest possible value for P (A ∩ B) is 0 (i.e. A and B can be disjoint).

(c) It is not possible for P (A ∩ B) = .7, since A ∩ B ⊂ A and P (A) = .6.

(d) The largest possible value for P (A ∩ B) is .3 (i.e. B can be contained in A, so A ∩ B = B is possible.)

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}.

Additional Problem

(2) For large numbers of runs (say ≥ 10000), the fraction \frac{wins}{total games} 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{2}{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.

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}

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} .)