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 Dc). There many ways of distributed events D in five places: (DDDcDDc), (DcDcDDD), etc. We can determine the number of possible combination using the combinatorial number 5C3.
Now the probability of P(DDDcDDc) = P(D)P(D)P(Dc)P(D)P(Dc) = (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 Dc occur. However, there are 5C3 = 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(Ac), 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.