In a box containing 60 bulbs, 6 are defective. What is the probability that out of a sample of 5 bulbs(i) none is defective(ii) exactly 2 are defective

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In a box containing 60 bulbs, 6 are defective. What is the probability that out of a sample of 5 bulbs
(i) none is defective
(ii) exactly 2 are defective

In a box containing 60 bulbs, 6 are defective. What is the probability that out of a sample of 5 bulbs
(i) none is defective
(ii) exactly 2 are defective

(i) Using Bernoulli’s Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$
$x=0,1,2, \ldots \ldots . . n \text { and } q=(1-p), n=5$
The probability of success, i.e. the bulb is defective $=p=\frac{6}{60}=\frac{1}{10}$
$q=1-\frac{1}{10}=\frac{9}{10}$
probability of that no bulb is defective piece=
$P(0$ defective items $)=$
$\begin{array}{l} { }^{5} C_{0} \cdot\left(\frac{1}{10}\right)^{0}\left(\frac{9}{10}\right)^{5} \\ \Rightarrow\left(\left(\frac{9}{10}\right)^{5}\right) \end{array}$
(ii) Using Bernoulli’s Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . .$ and $q=(1-p), n=5$
The probability of success, i.e. the bulb is defective $=p=\frac{6}{60}=\frac{1}{10}$
$\mathrm{q}=1-\frac{1}{10}=\frac{9}{10}$
probability of that there are exactly 2 defective pieces=
$\mathrm{P}(2$ defective items $)=$
$\begin{array}{l} { }^{5} C_{2} \cdot\left(\frac{1}{10}\right)^{2}\left(\frac{9}{10}\right)^{3} \\ \Rightarrow\left(\left(\frac{729}{10000}\right)\right) \end{array}$