A1. The probability that each item coming off a production line is defective is pand theprobability that it is non-defective is q, 0 < p < 1, p+q= 1. At the beginning of aday’s production, a quality control officer repeatedly inspects item each coming off aproduction line until he inspects nitems. Let Xbe the number of defective items hefinds.(a) Write down without proof the probability that X=k, indicating the possiblevalues of k. Hence, by considering the expansion of p+qn+q−pn. Show thatthe probability that Xis even is121 + (1 −2p)n
Question
A1. The probability that each item coming off a production line is defective is pand theprobability that it is non-defective is q, 0 < p < 1, p+q= 1. At the beginning of aday’s production, a quality control officer repeatedly inspects item each coming off aproduction line until he inspects nitems. Let Xbe the number of defective items hefinds.(a) Write down without proof the probability that X=k, indicating the possiblevalues of k. Hence, by considering the expansion of p+qn+q−pn. Show thatthe probability that Xis even is121 + (1 −2p)n
Solution
The problem is asking for the probability distribution of a random variable X, which represents the number of defective items found in n inspections. This is a binomial distribution problem, where each inspection is a Bernoulli trial with two possible outcomes: defective (with probability p) or non-defective (with probability q), and the trials are independent.
(a) The probability that X=k is given by the binomial probability formula:
P(X=k) = C(n, k) * p^k * q^(n-k)
where C(n, k) is the binomial coefficient, which gives the number of ways to choose k defective items from n inspections. The possible values of k are 0, 1, 2, ..., n.
To show that the probability that X is even is 1/2 * [1 + (1 - 2p)^n], we need to consider the expansion of (p+q)^(n) + (q-p)^(n).
The binomial theorem tells us that (p+q)^(n) = Σ [C(n, k) * p^k * q^(n-k)] for k=0 to n. This sum includes the probabilities of all possible outcomes, both even and odd.
Similarly, (q-p)^(n) = Σ [C(n, k) * (-p)^k * q^(n-k)] for k=0 to n. This sum also includes the probabilities of all possible outcomes, but the sign alternates between positive and negative for each term.
Adding these two expansions together, the terms for odd k will cancel out, leaving only the terms for even k. Dividing by 2 gives the average probability for even k, which is 1/2 * [1 + (1 - 2p)^n].
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