Question

• A key and difficult feature on a part in a job shop is inspected in batches of 20 for functionality and each part is classified as functional or non-functional. Historically, the non-functional rate is 4%, what is the probability that more than 2 parts in a batch of 20 are defective?

Answer 1

Let X denote the random variable representing the number of defective parts in a batch of 20.

Now, since the number of parts in a batch is fixed, each part has two outcomes (defective or non-defective), each part is defective with probability p = 4% = 0.04 independent of other parts. Thus, we can conclude that:
X ~ Bin(n = 20, p = 0.04) and the probability mass function of X is given by:

$\begin{align*} P(X=x ) &= \binom{n}{x}p^x(1-p)^{n-x} \ \ \ \ \ &&; x = 0,1,2,...,n \\ &= \binom{20}{x}*0.04^x*0.96^{20-x} \ \ \ \ \ &&; x=0,1,2,...,20 \end{align*}$

The probability that more than 2 parts are defective in a batch of 20 is given by:

$\begin{align*} \boldsymbol{P(X>2 )} &= 1 - P(X\le 2) \\ &= 1 - \left[P(X=0) + P(X=1) + P(X=2) \right ] \\ &= 1 - \left[\binom{20}{0}*0.04^0*0.96^{20} + \binom{20}{1}*0.04^1*0.96^{19} \right . \\ & \ \ \ \ \left .+ \binom{20}{2}*0.04^2*0.96^{18} \right ] \\ &= 1 - \left[0.442002 + 0.368335 + 0.145799 \right ] \\ &= 1 - 0.956137 \\ &= \bf 0.043863 \ \ \ \ \ \ \ \ [ANSWER] \end{align*}$

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