Question

A company manufacturers video games with a current defect rate of $0.95\%$. To make sure as few defective video games are delivered as possible, they are tested before delivery. The test is $98\%$ accurate at determining if a video game is defective. If $100000$ products are manufactured and delivered in a month, approximately how many defective products are expected to be delivered?

Answer 1

Finding the number of defective products:

Step-1: Construction of a Binomial Random Variable:

For each piece of product delivered, there are only two possible outcomes: either it is defective or it is not.

The probability of a piece of product delivered being defective is independent of any other pieces, which means that the binomial distribution is used to solve this problem.

Let $X$ be the random variable that denotes the number of defective product delivered. Here, we have to find expectation of $X$, $E\left(X\right)$.

Step-2: Finding the expectation:

Formula to be used: We know that for a binomial random variable $X$ with probability of success $p$ and with $n$ trials, the expectation is $E\left(X\right)=np$.

Here, $n=100000$.

Given that $0.95\%$ of the products are defective and the accuracy of the test before delivery is $98\%$. So, $(100-98)=2\%$ of the defective products are delivered. Hence, we must have:

$$\begin{gathered} p=0.95\%\times 2\% \\ =\frac{0.95}{100}\times \frac{2}{100} \\ =\frac{1.9}{10000} \end{gathered}$$

Hence,

$$\begin{gathered} E\left(X\right)=np \\ =100000\times \frac{1.9}{10000} \\ =19 \end{gathered}$$

Therefore, approximately $19$ defective products are expected to be delivered.