P (win a pizza) = P (A or B) = P (A) + P (B) – P (A and B) = .
Because the numbers of men and women in the school are about equal (that is, P (women) = 0.5), let
an even number represent a female and an odd number represent a male. Begin on the first line of the
table and consider groups of 12 digits. Count the even numbers among the 12. This will be the number
of females among the group. Repeat five times. The relevant part of the table is shown below, with
even numbers underlined and groups of 12 separated by two slanted bars (\):
In the five groups of 12 people, there were 3, 6, 3, 8, and 6 women. (Note: The result will, of course,
vary if a different assignment of digits is made. For example, if you let the digits 0 – 4 represent a
female and 5 – 9 represent a male, there would be 4, 7, 7, 5, and 7 women in the five groups.) So, in
40% of the trials there were 4 or fewer women in the class even though we would expect the average
to be 6 (the average of these 5 trials is 5.2). Hence, it seems that getting only 4 women in a class
when we expect 6 really isn’t too unusual because it occurs 40% of the time. (It is shown in the next
chapter that the theoretical probability of getting 4 or fewer women in a group of 12 people, assuming
that men and women are equally likely, is about 0.19.)
Because P (girl) = 0.6, let the random digits 1, 2, 3, 4, 5, 6 represent the birth of a girl and 0, 7, 8, 9
represent the birth of a boy. Start on the second row of the random digit table and move across the
line until you find the third digit that represents a girl. Note the number of digits needed to get three
successes. Repeat 5 times and compute the average. The simulation is shown below (each success,
i.e., girl, is underlined and separate trials are delineated by \).
It took 4, 7, 3, 6, and 5 children before they got their three girls. The average wait was 5. (The
theoretical average is exactly 5—we got lucky this time!) As with Exercise 15, the result will vary
with different assignment of random digits.
(a) P (x ≤ 22) = P (x = 20) + P (x = 21) + P (x = 22) = 0.2 + 0.3 + 0.2 = 0.7.
(b) P (x > 21) = P (x = 22) + P (x = 23) + P (x = 24) = 0.2 + 0.1 + 0.2 = 0.5.
(c) P (21 ≤ x < 24) = P (x = 21) + P (x = 22) + P (x = 23) = 0.3 + 0.2 + 0.1 = 0.6.
(d) P (x ≤ 21 or x > 23) = P (x = 20) + P (x = 21) + P (x = 24) = 0.2 + 0.3 + 0.2 = 0.7.
- (a) 18 of the 38 slots are winners, so P (win if bet on red)
(b) The probability distribution for this game is