Step 1

Introduction:

Denote X as the random variable of interest here, which is the number of defective items in the random sample of 10 items.

The batch from which the 10 items are randomly selected, is a large batch, which has 5% defectives. Due to the large size of the batch, it can be assumed that the selection of the 10 items are practically independent of each other, with probability of “success” or probability that each item is defective is \(\displaystyle{p}={0.05}\ {\quad\text{or}\quad}\ {5}\%\).

Thus, it may be assumed that, X has a binomial distribution with parameters, \(\displaystyle{n}={10}\) packages, \(\displaystyle{p}={0.05}\).

Since p denotes the probability of success, the probability of failure is, \(\displaystyle{q}={1}-{p}={0.95}\).

Step 2

Calculation:

If \(\displaystyle{X}\sim{B}in{o}{m}{i}{a}{l}{\left({n},{p}\right)}\), then the probability mass function of X is:

\(f(x)=\begin{cases}(\begin{array}{c}n\\x\end{array})p^{x}q^{n-x};x=0,1,...,n;\& 0\end{cases}\) Here, n is the number of independent trials, p is the probability of success in each trial, and q is the probability of failure.

Thus, the probability mass function of X here is:

\[f(x)=\begin{cases}(\begin{array}{c}10\\x\end{array})(0.05)^{x}(0.95)^{10-x};& x=0,1,2,...,10\\0; & otherwise\end{cases}\]

The entire batch, that is, 100% will be inspected if 2 or more defectives are found in the random sample of 10 items from that batch, that is, if \(\displaystyle{X}\geq{2}\). For a batch that meets the contract requirements (that is, a batch with 5% defectives), the probability that it will end up being 100% inspected is calculated below:

\(\displaystyle{P}{\left({X}\geq{2}\right)}={1}-{P}{\left({X}{ < }{2}\right)}\)

\(\displaystyle={1}-{\left[{P}{\left({X}\le{1}\right)}\right]}\)

\(\displaystyle={1}-{\left[{P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}\right]}\)

\(\displaystyle={1}-{\left[{f{{\left({0}\right)}}}+{f{{\left({1}\right)}}}\right]}\)

\(=1-\begin{cases}(\begin{array}{c}10\\0\end{array})(0.05)^{0}(0.95)^{10-0}+\begin{cases}(\begin{array}{c}10\\1\end{array})\end{cases}\end{cases}\)(0.05)^{1}(0.95)^{10-1}]\]

\(\displaystyle={1}-{\left[{\left({1}\right)}\cdot{\left({1}\right)}\cdot{\left({0.95}\right)}^{{{10}}}+{\left({10}\right)}\cdot{\left({0.05}\right)}\cdot{\left({0.95}\right)}^{{{9}}}\right]}\)

\(\displaystyle\approx{1}-{\left[{0.599}+{0.315}\right]}\)

\(\displaystyle={1}-{0.914}\)

\(\displaystyle={0.086}\).

Thus, the probability that the batch will end up being 100% inspected is Option A) 0.086.