A factory makes lamps. The probability that a lamp is defective is 0.05. A random sample of 30 lamps is tested.

Find the probability that there is at least one defective lamp in the sample.

A factory makes lamps. The probability that a lamp is defective is 0.05. A random sample of 30 lamps is tested.

Given that there is at least one defective lamp in the sample, find the probability that there are at most two defective lamps.

evidence of recognizing binomial (seen anywhere)     (M1)

e.g. \({\rm{B}}(n{\text{, }}p)\), \({0.95^{30}}\)

finding \({\rm{P}}(X = 0) = 0.21463876\)     (A1)

appropriate approach     (M1)

e.g. complement, summing probabilities

\(0.785361\)

probability is \(0.785\)     A1     N3

[4 marks]

identifying correct outcomes (seen anywhere)     (A1)

e.g. \({\rm{P}}(X = 1) + {\rm{P}}(X = 2)\) , 1 or 2 defective, \(0.3389 \ldots  + 0.2586 \ldots \)

recognizing conditional probability (seen anywhere)     R1

e.g. \({\rm{P}}(A|B)\) , \({\rm{P}}(X \le 2|X \ge 1)\) , P(at most 2|at least 1)

appropriate approach involving conditional probability     (M1)

e.g. \(\frac{{{\rm{P}}(X = 1) + {\rm{P}}(X = 2)}}{{{\rm{P}}(X \ge 1)}}\) , \(\frac{{0.3389 \ldots  + 0.2586 \ldots }}{{0.785 \ldots }}\) , \(\frac{{1{\text{ or }}2}}{{0.785}}\)

\(0.760847\)

probability is \(0.761\)     A1     N2

[4 marks]

Although candidates seemed more confident in attempting binomial probabilities than in previous years, some of them failed to recognize the binomial nature of the question in part (a). Many knew that the complement was required, but often used \(1 - {\rm{P}}(X = 1)\) or  \(1 - {\rm{P}}(X \le 1)\) instead of \(1 - {\rm{P}}(X = 0)\) .

Part (b) was poorly answered. While some candidates recognized that it was a conditional probability, very few were able to correctly apply the formula, identify the outcomes and follow on to achieve the correct result.

Only a few could find the intersection of the events correctly. Several thought the numerator was a product (i.e. \({\rm{P}}({\text{at most 2}}) \times {\rm{P({\text{at least 1}})}}\)), and then cancelled common factors with the denominator. Others realized that \(x = 1\) and \(x = 2\) were required but multiplied their probabilities.

This was the most commonly missed out question from Section A.