ProbabilityHardBloom L3

Question

The probability that an item produced by a factory is defective is ' p '. From a certain lot, a sample of ' $n$ ' items is drawn with replacement. If it contains no defective items, the lot is accepted, while if it has more than two defective items, the lot is rejected. If the sample has one or two defective items, an independent sample of ' $m$ ' items is drawn with replacement from the lot and combined with previous sample. If the combined sample does not contain more than two defective items, the batch is accepted. Find the probability that the batch is accepted.

Options

A.$(1 - p)^{n} + np(1 - p)^{n + m}$
B.$(1 - p)^{n} + np(1 - p)^{n + m - 1}$
C.$\left. \ (1 - p)^{n} + np(1 - p)^{n + m - 2}(1 + (m - 1)p) \right) + \ ^{n}C_{2}(1 - p)^{m + n - 2}p^{2}$
D.$(1 - p)^{n - 2}\left( (1 - p)^{2} + \frac{n(n - 1)p^{2}}{2} \right) + np(1 - p)^{n + m - 2}(1 + (m - 1)p)$

Solution

Sol. [Key Concept: Batch accepted if the combined sample has $\leq 2$ defectives — sum the binomial cases.]

Let q = 1 − p.

The batch is accepted in the following cases:

Case 1: The first sample contains no defective item.

P₁ = qⁿ

Case 2: The first sample contains exactly 1 defective item, and the second sample contains at most 1 defective item.

P₂ = npqⁿ⁻¹(qᵐ + mpqᵐ⁻¹)

Case 3: The first sample contains exactly 2 defective items, and the second sample contains no defective item.

P₃ = ⁿC₂p²qⁿ⁻²qᵐ

Therefore,

P = qⁿ + npqⁿ⁻¹(qᵐ + mpqᵐ⁻¹) + ⁿC₂p²qⁿ⁺ᵐ⁻²

Since q + mp = 1 + (m − 1)p,

we get P = qⁿ + npqⁿ⁺ᵐ⁻²[1 + (m − 1)p] + ⁿC₂p²qⁿ⁺ᵐ⁻²

Substituting q = 1 − p,

P = (1 − p)ⁿ + np(1 − p)ⁿ⁺ᵐ⁻²[1 + (m − 1)p] + ⁿC₂(1 − p)ⁿ⁺ᵐ⁻²p²

Hence, Answer: (C)

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