On the TI-83/84, the solution is given by 1–binompdf(12,0.003,0).
The clear message here is that even though the probability of any one failure seems remote
(0.003), the probability of at least one failure (3.5%) is large enough to be worrisome.
Because you already have 2 of the 5 prizes, the probability that the next box contains a prize you
don’t have is 3/5 = 0.6. If n is the number of trials until the first success, then P (X = n ) = (0.6). (0.4)
n −1 .
(a) P (X = 1) = (0.6)(0.4)1−1 = (0.6)(1) = 0.6. On the TI-83/84 calculator, the answer can be found
by geometpdf(0.6,1).
(b) P (X = 3) = (0.6)(0.4)3−2 = 0.096. On the calculator: geometpdf(0.6,3).
(c) The average number of boxes you will have to buy before getting the third prize is
- 40(0.8) = 32 and 40(0.2) = 8. The rule we have given is that both np and n (1 – p ) must be greater
than 10 to use a normal approximation. However, as noted in earlier in this chapter, many texts allow
the approximation when np ≥ 5 and n (1 – p ) ≥ 5. Whether the normal approximation is valid or not
depends on the standard applied. Assuming that, in this case, the conditions necessary to do a normal
approximation are present, we have - If p = 0.70, then μ = 0.70 and . Thus, P ( ≤ 0.65) =
. Since there is a very small probability of getting a sample
proportion as small as 0.65 if the true proportion is really 0.70, it appears that the San Francisco Bay
Area may not be representative of the United States as a whole (that is, it is unlikely that we would
have obtained a value as small as 0.65 if the true value were 0.70).Solutions to Cumulative Review Problems
The sample space for this event is {HHH, HHT , HTH , THH , HTT, HTH, THH, TTT}. One way
to do this problem, using techniques developed in Chapter 9 , is to compute the probability of each
event. Let X = the count of heads. Then, for example (bold faced in the list above), P (X = 2) = (0.6)
(0.6)(0.4) + (0.6)(0.4)(0.6) + (0.4)(0.6)(0.6) = 3(0.6)^2 (0.4) = 0.432. Another way is to take
advantage of the techniques developed in this chapter (noting that the possible values of X are 0, 1, 2,
and 3):
P (X = 0) = (0.4)^3 = 0.064; = binompdf(3,0.6,1)= 0.288;= binompdf(3,0.6,2)= 0.432; and P(X = 3) = =binompdf(3,0.6,3)= 0.216. Either way, the probability distribution is then: