Probability

Solutions by Bracket

A First Course in Probability

Chapter 4—Problems

4. Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a woman (for instance, X = 1 if the top-ranked person is female). Find P X = i , i = 1, 2, 3, . . . , 8, 9, 10.

Let Ei be the event that the the ith scorer is female. Then the event X = i correspdonds to the cc

event E1 E2 · · · Ei . It follows that

cc

P X = i = P (E1 E2 · · · Ei )

.

c

c

c

c

c

= P (E1 )P (E2 |E1 ) · · · P (Ei |E1 · · · Ei−1 )

Thus we have

P X=i

i

1/

1

2

5/

2

18

5/

3

36

5/

4

84

5/

5

252

1/

6

252

0.

7, 8, 9, 10

12. In the game of Two-Finger Morra, 2 players show 1 or 2 ﬁngers and simultaneously guess the number of ﬁngers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the ﬁngers shown by him and his opponent. If both players guess correctly or if neither players guess correctly, then no money is exchanged. Consider a speciﬁed player and denote by X the amount of money he wins in a single game of Two-Finger Morra. a. If each player acts independently of the other, and if each player makes his choice of the number of ﬁngers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the 4 possibilities is equally likely, what are the possible values of X and what are their associated probabilities?

A given player can only win 0, ±2, ±3, or ±4 dollars. Consider two players A and B , and let X denote player A’s winnings. Let Aij denote the event that player A shows i ﬁngers and guesses j , and deﬁne Bij similarly for player B.

1

We have P X = 2 = P (A11 B12 ) = P (A11 )P (B12 ) = 1 · 1 = 16 , since we have assumed that 44

1

Aij and Bij are independent and that P (Aij ) = P (Bij ) = 4 . Similarly, we have P X = 3 = 1

1

1

P (A12 B22 ∪ A21 B11 ) = 16 + 16 = 1 and P X = 4 = P (A22 B21 ) = 16 . Note that the situation 8

1

is completely symmetric for player B, so the we have P X = −2 = P X = −4 = 16 and 1

P X = −3 = 1 . Finally, we have P X = 0 = 1 − P X = 0 = 1 − 1 = 2 . 8

2

b. Suppose that each player acts independently of the other. If each player decides to hold up the same number of ﬁngers that he guesses his opponent will hold up, and if each player is equally likely to hold up 1 or 2 ﬁngers, what are the possible values of X and their associated probabilities? Neither player can win any money in this scenario. If player A shows 1 ﬁnger and guesses B will show 1 ﬁnger, then A can only win if B shows 1 ﬁnger. But if B shows 1 ﬁnger, then B will guess that A will show 1 ﬁnger, and thus neither player will win. The same holds for when A shows 2 ﬁngers and guesses that B will show 2 ﬁngers. Thus, we have P X = 0 = 1.

1

Mathematical Systems

Probability

20. A gambling book recommends the following “winning strategy” for the game of roulette. It recommends 18

that the gambler bet $1 on red. If red appears (which has probability 38 ), then the gambler should take her $1 proﬁt and quit. If the gambler loses this bet (which has probability 20 of occurring), she should 38

make additional $1 bets on red on each of the next two spins of the roulette wheel and then quit. Let X denote the gambler’s winnings when she quits.

a. Find P X > 0 .

Note that X only takes on the values −3, −1, and 1. Thus P X>0 =P X=1

= P (she wins immediately or she loses and then wins the next two) = P (she wins immediately) + P (she loses and then wins the next two) 18 20 18 18

=

+

·

·

≈ .592

38 38 38 38

b. Are you convinced that the winning strategy is indeed a “winning” strategy? Explain your answer! The expected value of X is negative (≈ −.108), which is accounted for by the fact that although the gambler has a high probability of winning $1,...