Mathematics 375–Probability and Statistics I Solutions for Problem Set 3 September 23, 2005

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} .)

2.65/ P (A) < P (A|B) = \frac{P (A∩B)}{P (B)} implies that P (A)P (B) < P (A ∩ B) (note we are

given that P (B) > 0). Hence P (B) < \frac{P (A∩B)}{P (A)} = P (B|A).

2.67/ We use the characterization A, B independent ⇔ P (A ∩ B) = P (A)P (B). a) Since A = (A ∩ B) ∪ (A ∩ \overline{B}) and (A ∩ B) ∩ (A ∩ \overline{B}) = ∅, we have P (A) = P (A ∩ B) + P (A ∩ \overline{B}). Hence

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

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

b) The answer is yes: Similarly, 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 part a
= P (\overline{B})(1 − P (A))
= P (\overline{B})P (\overline{A})

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

2.70/ a) P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = .95 + .90 − .88 = .97. b) P (\overline{A ∪ B}) = 1 − .97 = .03.

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.

2.79/ a) From problem 2.40, there were 8 likely winning numbers out of the 1000 possible combinations of 3 digits 0, 1, …, 9. a) Assuming that the drawings in CT and PA were independent (note the “all indications CT was due to pure chance” in problem!), P (666 in CT | 666 in PA) = P (666 in CT) = 1/1000. b) Again by independence, P (666 in CT and 666 in PA)(1/8)(1/1000) = 1/8000.

2.88/ Let D_{i} be the event that the ith radar set detects the plane. P (D_{i}) = .98 and P (\overline{D_{i}}) = .02. a) Since D_{i} are independent P (\overline{D_{i}}) are also independent (see 2.67b above). Hence P (\overline{D_{1}} ∩ \overline{D_{2}} ∩ \overline{D_{3}}) = (.02)^{3} = .000008. b) The probability that all 3 radar sets do detect the plane is P (D_{1} ∩ D_{2} ∩ D_{3}) = (.98)^{3} = .941192

2.91/ Because of independence (.75)^{4}.

2.101/ By Bayes’ Rule,

P (I_{1}|H) = \frac{(.9)(.01)}{(.9)(.01) + (.95)(.005) + (.75)(.02)} ≐ .313

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).

2.103/ Let N denote a negative reaction, \overline{N} a positive one, M denote a male, \overline{M} a female (that means no mixed reactions, no hermaphrodites or neuters :) ). The given information in the problem is: P (N|\overline{M}) = .3, P (N|M) = .6, which implies P (\overline{N}|\overline{M}) = .7, P (\overline{N}|M) = .4. Finally P (M) = .25 and P (\overline{M}) = .75. Hence by Bayes’ Rule, P (M|N) = \frac{(.6)(.25)}{(.6)(.25)+(.3)(.75)} = .4.