There are $6 \%$ defective items in a large bulk of times. Find the probability that a sample of 8 items will include not more than one detective item.

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There are $6 \%$ defective items in a large bulk of times. Find the probability that a sample of 8 items will include not more than one detective item.

Using Bernoulli’s Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$
$x=0,1,2, \ldots \ldots . .$ and $q=(1-p), n=8$
The probability of success, i.e. the bulb is defective $=p=\frac{6}{100}=\frac{6}{100}$
$\mathrm{q}=1-\frac{6}{100}=\frac{94}{100}$
probability of that there is not more than one defective piece=
$P(0$ defective items $)+P(1$ defective item $)=$
$\begin{array}{l} { }^{8} C_{0} \cdot\left(\frac{6}{100}\right)^{0}\left(\frac{94}{100}\right)^{8}+{ }^{8} C_{1} \cdot\left(\frac{6}{100}\right)^{1}\left(\frac{94}{100}\right)^{7} \\ \Rightarrow\left(\left(\frac{47}{50}\right)^{7} \times\left(\frac{71}{50}\right)\right) \end{array}$