Question

A box contains $100$ bulbs, out of which $10$ are defective. A sample of $5$ bulbs is drawn. The probability that none is defective, is

• A. $\frac{9}{10}$
• B. ${\left(\frac{1}{10}\right)}^5$
• C. ${\left(\frac{9}{10}\right)}^5$
• D. ${\left(\frac{1}{2}\right)}^5$

Answer 1

The correct option is C

${\left(\frac{9}{10}\right)}^5$

Explanation for the correct answer:

Step- 1: Given data and variable :

Total number of bulbs = $100$

Number of bulbs defective = $5$

Let us consider, Number of defective bulbs be $X$.

So, selecting the bulb follows Bernoulli trial. Then $X$ has a binomial distribution.

Step- 2: Finding $\mathit{n}\mathbf{,}\mathit{p}\mathbf{,}\mathit{q}$:

Let $n$ be the number of times selecting the bulb, $n=5$

Let $p$ be the probability of getting defective bulbs, $p=\frac{10}{100}=\frac{1}{10}$

And $q$is probability of getting non-defective bulb

We know that,

$$\begin{gathered} q=1-p \\ =1-\frac{1}{10} \\ =\frac{9}{10} \end{gathered}$$

Step- 3: Finding the probability:

The formula for binomial distribution,

$\mathit{P}\mathbf{(}\mathbf{X}\mathbf{=}\mathbf{x}\mathbf{)}{\mathbf{=}}^{\mathbf{n}}{\mathit{C}}_{\mathbf{x}}\mathbf{}{\mathit{q}}^{\mathbf{n}\mathbf{-}\mathbf{x}}\mathbf{}{\mathit{p}}^{\mathbf{x}}$

The probability of none bulb is defective, $x=0$

On substituting the above values,

$$\begin{gathered} P(X=0)={}^5{C}_0{\left(\frac{9}{10}\right)}^{5-0}{\left(\frac{1}{10}\right)}^0 \\ =1\times {\left(\frac{9}{10}\right)}^5\times 1 \\ ={\left(\frac{9}{5}\right)}^5 \end{gathered}$$

Hence, option (C) is the correct answer