There are 5 % defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
There are 5 % defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Solution:

Let’s say there are x defective items in a sample of ten drawn sequentially. As we can see, the items’ drawings are now done with replacement. As a result, the trials are Bernoulli tests.

Now, probability of getting a defective item, p=5 / 100=1 / 20

Thus, q=1-1 / 20=19 / 20

therefore We can say that x has a binomial distribution, where n=10 and p=1 / 20

Thus,
\mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \mathrm{p}^{\mathrm{x}}, where \mathrm{x}=0,1,2 ldots \mathrm{n}

Probability of getting not more than one defective item =\mathrm{P}(\mathrm{X} leq 1)

=P(X=0)+P(X=1)

={ }^{10} \mathrm{C}{0}(19 / 20)^{10}(1 / 20)^{0}+{ }^{10} \mathrm{C}{1}(19 / 20)^{9}(1 / 20)^{1}

=\left(\frac{19}{20}\right)^{10}+10 \times\left(\frac{19}{20}\right)^{9}\left(\frac{1}{20}\right)^{1}
=\left(\frac{19}{20}\right)^{9}\left[\frac{19}{20}+\frac{10}{20}\right]
=\left(\frac{19}{20}\right)^{9} \times\left(\frac{29}{20}\right)