Outcome    -4     -2       0       2        4

Probability 1/16  4/16    6/16   4/16     1/16
         (0.0625) (0.25) (0.375) (0.25) (0.0625)

7.58. (a) The given probabilities have sum 0.96, so P(AB) = 0.04. (b) Use Rule C: P(not 0) = 1 - 0.49 = 0.51. (c) No - blood type is categorical, not quantitative.

7.59. If the deck is shuffled so that each card is equally likely to be dealt, each of the 13 face values has probability 1/13. There are many other legitimate assignments: say P(Ace) = 1 and all others have 0 probability. There are also many possible
illegitimate choices, such as P(Ace) = 0.6 and P(King) = 0.5 with all other values having 0 probability; this assigns total probability exceeding 1.

7.60. We see that Duke has probability 0.3 and North Carolina has probability 0.6; adding the 0.2 for the other schools named gives 1-J, so this is not a legitimate assignment.

7.61. (a) The numbers given are all between 0 and 1, so just check that their sum is 1. (b) This event is 19, 10, 11, 12}. Its probability is 0.931. (c) 11.251 years of school.

7.62. (a) The probability is about 1/50, or 0.02 (see Exercise 7.6).

(b) 0.015 = 3/200 so the odds are 197 to 3 (or 985 to 15, or 65.6 to 1, etc.).

7.63.

0, 1 - professional

2, 3, 4, 5, 6 - white collar

7, 8 - blue collar

9 - no steady occupation

(a) A single random digit simulates the son's class, as in the table above. (b) Five
digits simulate the five sons' occupations, so 250 digits are needed. The correct
probability, calculated from a binomial distribution with n = 5 and p = 0.2, is about 0.262.

7.64. (a) 0.69. (b) $10.

00 - wins $250

01 to 05 - wins $100

06 to 30 - wins $10

31 to 99 - wins nothing

(c) Use two digits to simulate one customer, as in the table above. (d) They play
independently, so three successive two-digit groups simulate their winnings. Then add the three amounts won to get their pooled winnings. (e) The true expected total is $30, the sum of their individual expected values. They can win at least $100 only if at least one of them wins at least $100 (even three $10 prizes won't do it). Each has probability 0.06 of winning at least $100 and probability 0.94 of failing. So the true value of the probability to be simulated is 1 - [(0.94)raised to the third power] or 0. 169.

7.65. The birthday of a single person is simulated by a three-digit group of random
digits. Groups 001 to 365 stand for the days of the year (ignoring those born on February 29) , and others are ignored. By the second assumption of the model
(independence} 23 consecutive three-digit groups (skipping those not 001 to 365) simulate the birth dates of 23 persons. Line 139 gives 356, 356, 1717 ... , so that the first two persons have the same birthday! (Purely by chance, not by design.)

The birthday problem is discussed in many texts on probability. With 23 persons, the probability of at least one match is 0.507, computed using Rule C for probabilities as

1 - [(365/365) (364/365) (363/365) ... (344/365) (343/365)]
If you try this in class, you should make sure you have no twins (or triplets, etc.), which would violate the independence assumption.


7.66. The probability model is

Outcome     2    3    4    5    6    7    8    9   10   11   12
Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

and from this the expected value is 252/36 = 7.

7.67. (a) The probability model is shown in the table below;
the expected value is therefore -$0.25.

Winnings     $2   -$1
Probability 0.25 0.75

(b) Using the probability given, the expected value is
(0.06)(11) + (0.94)(-1) = -$0.28, so this bet is slightly less favorable.

7.68. The probability of a gain should be given as 1/2 on any day. Each day is simulated by a single random digit with (say) odd = up and even = down. Consecutive digits simulate successive days; stop as soon as the number of ups exceeds the number of downs, or after 5 days in any event. The correct probability model is

Outcome       +1     -1     -3     -5

Probability 22/32   5/32   4/32   1/32

and so the true expected value is exactly 0! This is an example of the general fact mentioned in the text: There is no winning system for multiple plays of a fair game.