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**Answer**

Binomial Distribution problems worksheet

**Example 1**

It is expected that 10% of production from a continous process will be defective.

Find the probability that in a sample of 10 units chosen at random exactly 2 will be defective and atleast 2 will be defective.

Solution

Exactly 2 will be defective;

$P(X=2)=\left(\begin{array}{c}10\\ 2\end{array}\right){(0.1)}^{2}{(0.9)}^{10-2}$

$=\left(\begin{array}{c}10\\ 2\end{array}\right){(0.1)}^{2}{(0.9)}^{8}$

$=45\times 0.01\times 0.43047\phantom{\rule{0ex}{0ex}}=0.1937$

Atleast 2 will be defective

$P(x\ge 2)=P(x=2)+P(x=3)+........P(x=10)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$=1-P(x<2)$

$P(X\ge 2)=1-\{\left(\begin{array}{c}10\\ 0\end{array}\right){(0.1)}^{0}{(0.9)}^{10}+\left(\begin{array}{c}10\\ 1\end{array}\right){(0.1)}^{1}{(0.9)}^{9}\}$

$=1-(0.3487+0.3874)$

=0.2640

**Example 2**

Probability that a computer drawn from a batch of computer its defective is 0.1. If a sample of 6 computer is taken. Find the probability that it will contain;

(i) No defective computer

$P(X=0)=\left(\begin{array}{c}6\\ 0\end{array}\right){(0.1)}^{0}{(0.9)}^{10}$

$\phantom{\rule{0ex}{0ex}}=1\times 1\times 0.34867844$

=0.3487

(ii) 5 or 6 are defective Computer

$P(X<5)=\left(\begin{array}{c}6\\ 5\end{array}\right){(0.1)}^{5}{(0.9)}^{1}+\left(\begin{array}{c}6\\ 6\end{array}\right){(0.1)}^{6}{(0.9)}^{0}$

=0.000054+0.000001

=0.000055

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