Solutions to Assigned Exercises from Chapter 7 (Errors may have been introduced by the scanning process.)

Section 1: What is probability?

7.7. (a) 0. (b) 1. (c) 0.01. (d) 0.6 (or 0.99, but "more often than not" is a rather weak description of an event with probability 0.99!)

7.9. (a) The two given probabilities add to 0.73, so the remaining probability must be 0.27 (assuming that there were no "no opinion" responses). (b) 0.73.

Section 2: Thinking about probability

7.20. (a) When choosing four digits -for the lottery (or from the table of random digits), any of the 10,000 possible combinations (0000, 0001, .. . , 9999) will show UP with the same (long-run) frequency. Like a coin, the pick-four machine has "no memory." (b) Most people would mistakenly think that 2873 is more likely than 9999 - the latter does not "look random" (the myth of short-run regularity).

7.22. After 6 tosses: 0. After 100 tosses: 47/100 = 0.47. After 1000 tosses: 497/1000 = 0.497. After 10,000 tosses: 4997/10000 = 0.4997.

7.24. Weather in successive winters is probably less than completely independent (consider the Ice Age, for example); the "law of averages" is no more reliable for forecasting the weather (in the short run) than it is for other predictions.

7.25. (a) The wheel is not affected by its past outcomes - it has no memory; outcomes are independent. So on any one spin, black and red remain equally likely. (b) Removing a card changes the composition of the remaining deck, so successive draws are not independent. If you hold 5 red cards, the deck now contains 5 fewer red cards, so your chance of another red decreases.

7.26. (a) The relative frequency is more likely to be close to 1/2 in a longer series of tosses, so choose 100 tosses. [In 10 tosses, the calculated probability of 4, 5, or 6 heads is 0.65625 from the binomial distribution. In 100 tosses, the probability of 40 to 60 heads (inclusive) is about 0.9648 - this can also be approximated (as 0.964) using the normal approximation with continuity correction.] (b) It becomes less likely that exactly half will be heads as the number of tosses increases. [The calculated probability of 5 heads in 10 tosses is 0.246. The probability of 50 heads in 100 tosses is about 0.0796. An even more extreme example: the probability of 1 head in 2 tosses is 0.5.]

7.33. (a) A single hole succeeds with probability 0.1 and fails with probability 0.9. Successive holes are independent. (b) A single digit simulates drilling a hole, with 0 a hit and 1 to 9 a dry hole. Ten successive digits simulate 10 holes; if we wish, we can stop picking digits as soon as oil is struck without. changing the probability of going broke. (c) Starting in line 140 and inspecting successive groups of ten digits for O's gives 12 successes in 20 tries. (It turns out that the answer is the same if we stop picking digits after a success, but this is a more tedious procedure.) So the estimated probability of going broke is 8/20 = 0.4. [The true, calculated probability is (0.9) to the tenth power = 0.349.]

7.39.

0, 1, 2, 3 - no female offspring

4, 5, 6, 7 - one female offspring

8, 9 two female offspring

(a) A single digit simulates the fecundity of a single female as shown above.

(b) The result of the simulation varies with the portion of the table employed, but generally there will be markedly fewer females in the fifth
generation, so that the population is clearly dying out. [The expected number of
females born to a given female is 0.8; that is, we "expect" a 20% reduction in the
number of females from one generation to the next. See Exercise 7.46 in the next
section.]


7.43. (a) The expected value is (600) (1/2) + (0)(1/2) = 300 people. (b) There is no
difference (except in the phrasing): saving 400 is the same as losing 200. (c) No - the choice seems to be based on how the options "sound."

7.46. (a) For the Asian stochastic beetle: (0)(0.2) + (1)(0-3) + (2)(0.5) = 1.3.

(b) For the benign boiler beetle: (0)(0.4) + (1)(0.4) + (2)(0.2) = 0.8. (c) When a large population of beetles is considered, each generation of Asian stochastic beetles will contain close to 1.3 times as many females as the preceding generation. So the population will grow steadily. Each generation of benign boiler beetles, on the other hand, contains only about 80% as many females as the preceding generation.

7.51. Taking male and female children to be
equally likely, the expected profit
is ($0)(1/2) + ($10)(1/2) = $5.

(In fact, males are very slightly more common, so the psychic should guess 'boy')