Question

A bag contains $2$ white and $4$ black balls. A ball is drawn $5$ times with replacement. The probability that at least $4$ of them are white is:
(A) $\dfrac{8}{{141}}$
(B) $\dfrac{{10}}{{243}}$
(C) $\dfrac{{11}}{{243}}$
(D) $\dfrac{8}{{41}}$

Answer 1

Hint: In the given question we are provided with a bag containing some black and white balls. We are required to find the probability of picking at least four white balls when a ball is drawn five times from the bag at random with replacement. So, we will first find the probability of picking white ball and black ball from the bag. Then, we will find the probability of having four and five white balls and add them up so that we get the total probability of at least four white balls. We will also make use of the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Complete answer:
Number of white balls $ = 2$
Number of black balls $ = 4$
So, total number of balls in bag $ = 2 + 4 = 6$
Now, probability of drawing a white ball $ = \dfrac{{Favourable\,number\,of\,outcomes}}{{Total\,number\,of\,outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$
Probability of drawing black ball $ = \dfrac{{Favourable\,number\,of\,outcomes}}{{Total\,number\,of\,outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}$
Now, we have to find the probability that at least four of the five balls selected are white.
Let X be the number of balls. Then, $P\left( {X \geqslant 4} \right) = P\left( {X = 4} \right) + P\left( {X = 5} \right)$.
So, we find the probability of drawing four white balls.
Now, any four of the five balls can be white. So, the number of ways of selecting four out of five balls is $^5{C_4} = 5$.
Now, the probability of drawing four white and one black ball \[{ = ^5}{C_4} \times {\left( {\dfrac{2}{6}} \right)^4} \times \left( {\dfrac{4}{6}} \right)\]
\[ = 5 \times {\left( {\dfrac{1}{3}} \right)^4} \times \left( {\dfrac{2}{3}} \right)\]
Doing the calculations, we get,
\[ = \dfrac{{10}}{{{3^5}}} = \dfrac{{10}}{{243}}\]
Now, we will find the probability of drawing five white balls.
So, five of the five balls must be white. So, the number of ways of selecting five out of five balls is $^5{C_5}$.
Now, the probability of drawing five white balls \[{ = ^5}{C_5} \times {\left( {\dfrac{2}{6}} \right)^5} \times {\left( {\dfrac{4}{6}} \right)^0}\]
\[ = 1 \times {\left( {\dfrac{1}{3}} \right)^5} \times {\left( {\dfrac{2}{3}} \right)^0}\]
Doing the calculations, we get,
\[ = \dfrac{1}{{{3^5}}} = \dfrac{1}{{243}}\]
Now, we know that $P\left( {X \geqslant 4} \right) = P\left( {X = 4} \right) + P\left( {X = 5} \right)$. So, the probability of drawing at least four white balls is the sum of the two probabilities.
So, probability of drawing at least four white balls \[ = \dfrac{{10}}{{243}} + \dfrac{1}{{243}} = \dfrac{{11}}{{243}}\].
Hence, option (C) is correct.

Note:
The question revolves around the concepts of probability and permutations and combinations. One should know about the principle rule of counting. Care should be taken while handling the calculations. Calculations should be verified once so as to be sure of the answer. We must know the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to calculate the number of ways of selecting r things out of n different things.