# There are 5% defective items in a large bulk of items.

Question:

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 X denote the number of defective items in a sample of 10 items drawn successively. Since the drawing is done with replacement, the trials are Bernoulli trials.

$\Rightarrow p=\frac{5}{100}=\frac{1}{20}$

$\therefore q=1-\frac{1}{20}=\frac{19}{20}$

$X$ has a binomial distribution with $n=10$ and $p=\frac{1}{20}$

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

$={ }^{19} \mathrm{C}_{x}\left(\frac{19}{20}\right)^{10-x} \cdot\left(\frac{1}{20}\right)^{x}$

P (not more than 1 defective item) = P (X ≤ 1)

$=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)$

$={ }^{10} \mathrm{C}_{0}\left(\frac{19}{20}\right)^{10} \cdot\left(\frac{1}{20}\right)^{0}+{ }^{10} \mathrm{C}_{1}\left(\frac{19}{20}\right)^{9} \cdot\left(\frac{1}{20}\right)^{1}$

$=\left(\frac{19}{20}\right)^{10}+10\left(\frac{19}{20}\right)^{9} \cdot\left(\frac{1}{20}\right)$

$=\left(\frac{19}{20}\right)^{9} \cdot\left[\frac{19}{20}+\frac{10}{20}\right]$

$=\left(\frac{19}{20}\right)^{9} \cdot\left(\frac{29}{20}\right)$

$=\left(\frac{29}{20}\right) \cdot\left(\frac{19}{20}\right)^{4}$

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