Question

A box contains $6$ pens, $2$ of which are defective. Two pens are taken randomly from the box. If random variable. X: Number of defective pens obtained, then standard deviation of X =
(A) $\pm \dfrac{4}{{3\sqrt 5 }}$
(B) $\dfrac{8}{3}$
(C) $\dfrac{{16}}{{45}}$
(D) $\dfrac{4}{{3\sqrt 5 }}$

Answer 1

It is given that there is a box that contains pens and there are some defective pens also. We have to find the standard deviation for the number of defective pens obtained. First we need to find the probabilities of getting the defective pen then putting it in the above formula we will get the required result.

Formula used: We know, if the generator of random variable X is discrete with probability mass function ${x_1} \to {p_1},{x_2} \to {p_2},{x_3} \to {p_3},.....,{x_n} \to {p_n}$, then
$Var\left( X \right) = \left( {\sum\limits_{i = 1}^n {{p_i}{x_i}^2} } \right) - {\mu ^2}$
Where $\mu = \sum\limits_{i = 1}^n {{p_i}{x_i}}$ is the expected value.
The formula of Standard deviation is,
$\sigma = \sqrt {Var\left( X \right)}$
Combination:
In mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter.
For a combination,
$C\left( {n,r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Where, factorial n is denoted by $n!$ and defined by
$n! = n(n - 1)(n - 2)(n - 4).......2.1$

Complete step-by-step solution:
It is given that a box contains $6$ pens, $2$ of which are defective. Two pens are taken randomly from the box.
Let the random variable X represent the number of defective pens obtained.
Then X can be $0,1,2$.
We need to find out the correct option which is the standard deviation of X.
The probability of getting zero defective pen is $P(X = 0) = \dfrac{{^4{C_2}}}{{^6{C_2}}}$
Using the combination formula,
$\Rightarrow \dfrac{{\dfrac{{4!}}{{2!2!}}}}{{\dfrac{{6!}}{{2!4!}}}}$
Expanding the factorial terms,
$\Rightarrow \dfrac{{\dfrac{{4 \times 3 \times 2}}{{2 \times 2}}}}{{\dfrac{{6 \times 5 \times 4!}}{{2 \times 4!}}}}$
Hence we get,
$\Rightarrow \dfrac{6}{{15}}$
The probability of getting one defective pen is $P(X = 1) = \dfrac{{^2{C_1}{ \times ^4}{C_1}}}{{^6{C_2}}}$
By using the combination formula,
$\Rightarrow \dfrac{{\dfrac{{2!}}{{1!1!}} \times \dfrac{{4!}}{{1!3!}}}}{{\dfrac{{6!}}{{2!4!}}}}$
Expanding the factorial terms,
$\Rightarrow \dfrac{{2 \times \dfrac{{4 \times 3!}}{{3!}}}}{{\dfrac{{6 \times 5 \times 4!}}{{2 \times 4!}}}}$
Hence, we get,
$\Rightarrow \dfrac{8}{{15}}$
The probability of getting two defective pens= $P(X = 2) = \dfrac{{^2{C_2}}}{{^6{C_2}}}$
By using the combination formula,
$\Rightarrow \dfrac{{\dfrac{{2!}}{{2!1!}}}}{{\dfrac{{6!}}{{2!4!}}}}$
Expanding the factorial terms,
$\Rightarrow \dfrac{1}{{\dfrac{{6 \times 5 \times 4!}}{{2 \times 4!}}}}$
Hence we get,
$\Rightarrow \dfrac{1}{{15}}$
Therefore putting all the values in variance formula we get $Var\left( X \right) = \left( {\dfrac{6}{{15}} \times {0^2} + \dfrac{8}{{15}} \times {1^2} + \dfrac{1}{{15}} \times {2^2}} \right) - {\left( \mu \right)^2}$
Where, $\mu = \dfrac{6}{{15}} \times 0 + \dfrac{8}{{15}} \times 1 + \dfrac{1}{{15}} \times 2$
$\Rightarrow \dfrac{8}{{15}} + \dfrac{2}{{15}}$
Simplifying we get,
$\Rightarrow \dfrac{{10}}{{15}} = \dfrac{2}{3}$
Thus,
$Var\left( X \right) = \left( {\dfrac{8}{{15}} + \dfrac{4}{{15}}} \right) - {\left( {\dfrac{2}{3}} \right)^2}$
$\Rightarrow \dfrac{{12}}{{15}} - \dfrac{4}{9}$
Simplifying we get,
$\Rightarrow \dfrac{{36 - 20}}{{45}} = \dfrac{{16}}{{45}}$
Hence the standard deviation of X is $\sigma = \sqrt {Var\left( X \right)} = \sqrt {\dfrac{{16}}{{45}}} = \pm \dfrac{4}{{3\sqrt 5 }}$

$\therefore$ Option (A) is the correct answer.

Note: Standard deviation measures the dispersion of a dataset relative to its mean. A volatile stock has a high standard deviation, while the deviation of a stable blue-chip stock is usually rather low. As a downside, the standard deviation calculates all uncertainty as risk, even when it’s in the investor’s favour such as above average returns.