Question

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If “X” be the number of white balls drawn, the $\dfrac{mean\text{ of X}}{standard\text{ deviation of X}}$ is equal to :-
(a). 4
(b). $\dfrac{4\sqrt{3}}{2}$
(c). $4\sqrt{3}$
(d).$3\sqrt{2}$

Answer 1

Hint: We will approach the problem by defining a random variable for this random experiment. If we see the problem very carefully, we are able to understand the experiment of picking a ball from the bag one by one with replacement is a random experiment and the experiment has only two outcomes.

Complete step-by-step answer:
Suppose, “X” denotes the number of white balls. The probability of getting one white ball is \[\dfrac{30}{40}=\dfrac{3}{4}\]and one red ball is \[\dfrac{10}{40}=\dfrac{1}{4}\].
We also assume that getting white ball is “success” and getting a red ball is a “failure”.
In probability theory this type of experiment is known as “Binomial Trial”.
In 1713 Swiss Mathematician Jakob Bernouli discovered this type of distribution. As there are two monomials added, it was named as a “Binomial Trial” and corresponding distribution is “Binomial Distribution”.

Here the probability distribution function is, that is the probability of X taking the value r is the r-th term of the binomial expansion of \[{{(p+q)}^{16}}\] where p=\[\dfrac{3}{4}\] and q= \[\dfrac{1}{4}\].
For the random variable “X”, the corresponding probability distribution function is
\[P\left( X=r \right){{=}^{16}}{{C}_{_{r}}}{{\left( \dfrac{3}{4} \right)}^{16-r}}{{\left( \dfrac{1}{4} \right)}^{r}}\]
Where r can take values between 0 to 16. Both inclusive.
Now we need two formulae of mean and standard deviation, for a random variable X
Mean of X , \[\left\langle X \right\rangle =\sum\limits_{r=0}^{\infty }{rP(X=r)}=np\]
Standard deviation of X ,\[\sqrt{\left\langle {{X}^{2}} \right\rangle -{{\left\langle X \right\rangle }^{2}}}=\sqrt{npq}\]
Where p=\[\dfrac{3}{4}\] and q= \[\dfrac{1}{4}\].

So, $\dfrac{mean\text{ of X}}{standard\text{ deviation of X}}$=$\dfrac{16.\dfrac{3}{4}}{\sqrt{16.\dfrac{3}{4}.\dfrac{1}{4}}}=\dfrac{12}{\sqrt{3}}=4\sqrt{3}$

So, the correct answer is “Option c”.

Note: If we do not know the concept of binomial distribution, then we can’t solve the problem.
Please don’t confuse yourself with the binomial distribution with “Bernoulli Trial”.