**23. Incorrect. The answer is
false not true.** Let's
represent the inspections by five spaces _ _ _ _ _. In each space we can place
the events D or not D (or D^{c}). There many ways of distributing
events D in five places: (DDD^{c}DD^{c}),
(D^{c}D^{c}DDD), etc. We can determine
the number of possible combination using the combinatorial number _{5}C_{3}.

Now the probability of P(DDD^{c}DD^{c})
= P(D)P(D)P(D^{c})P(D)P(D^{c}) = (1/6)(1/6)(5/6)(1/6)(5/6) =
(1/6)^{3}(5/6)^{2} because the problem assumes independence.
Similarly, we have that: P = (1/6)^{3}(5/6)^{2} for all other
outcomes in which three D and two D^{c} occur. However, there are _{5}C_{3}
= 10 such outcomes, and they are mutually exclusive. Therefore, the required probability is:

In general, as Schaum's outline shows, if p = P(A) and q = P(A^{c}), then by using the same
reasoning as given above, the probability of getting exactly 'x' A events in
'n' independent trials is:

This formula is also known as the *discrete
binomial distribution of Bernoulli*'s *probabilistic
distribution.*