**23. Correct. The answer is false.** Let's represent the inspections be represented
by five spaces _ _ _ _ _. In each space we can place the events D or not D (or
D^{c}). There many ways of distributed 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.*