Question

A bag contains $6$ red , $4$ white and $8$ blue balls . If three balls are drawn at random , then the probability that $2$ are white and $1$ is red is
A.\[\left( 1 \right)\]\[\dfrac{5}{{204}}\]
B.\[\left( 2 \right)\]\[\dfrac{7}{{102}}\]
C.\[\left( 3 \right)\]\[\dfrac{3}{{68}}\]
D.\[\left( 4 \right)\]\[\dfrac{1}{{13}}\]

Answer 1

Hint: We have to find the required probability . We solve this question using the concept of probability and also the concept of arrangements of the balls using permutation and combination . We firstly find the total possible arrangements and also the favourable outcomes using the formula of combination . The probability is given by favourable outcomes to the total possible arrangements.

Complete step-by-step answer:
Given :
Total number of balls \[ = {\text{ }}6{\text{ }} + {\text{ }}4{\text{ }} + {\text{ }}8\]
Total number of balls \[ = {\text{ }}18\]
Now ,
Number of balls drawn \[ = {\text{ }}3\]
As , we know the formula of combination
\[{}^n{C_r} = {\text{ }}\dfrac{{n!}}{{r!{\text{ }} \times {\text{ }}\left( {n{\text{ }} - {\text{ }}r} \right)!}}\]
Applying the formula , we get
The total ways of arrangements of the ball \[ = {\text{ }}{}^{18}{C_3}\]
The total ways of arrangements of the ball \[ = {\text{ }}\dfrac{{{\text{ }}18!{\text{ }}}}{{\left( {{\text{ }}3!{\text{ }} \times {\text{ }}15!{\text{ }}} \right)}}\]
The total ways of arrangements of the ball \[ = {\text{ }}816\]
Let a be the possible favourable outcomes
Now ,
As according to the question we need $2$ white and $1$ red ball . So the required favourable outcomes would be equal to the combination of white balls into the combination of red balls , which is given as
Possible favourable outcomes \[n\left( a \right){\text{ }} = {\text{ }}{}^4{C_2}{\text{ }} \times {\text{ }}{}^6{C_1}\]
Using the formula of combination
\[n\left( a \right){\text{ }} = {\text{ }}\left[ {\dfrac{{4!{\text{ }}}}{{2!{\text{ }} \times 2!}}} \right]{\text{ }} \times {\text{ }}\left[ {\dfrac{{6!{\text{ }}}}{{1!{\text{ }} \times {\text{ }}5!}}} \right]\]
\[\;n\left( a \right){\text{ }} = {\text{ }}6{\text{ }} \times {\text{ }}6\]
Further simplifying we get,
\[n\left( a \right){\text{ }} = {\text{ }}36\]
\[The{\text{ }}required{\text{ }}probability{\text{ }} = \dfrac{{total{\text{ }}favourable{\text{ }}outcomes}}{{total{\text{ }}possible{\text{ }}outcomes}}\]
Required probability \[ = \dfrac{{36}}{{816}}\]
Cancelling the terms , we get
equired probability \[ = {\text{ }}\dfrac{3}{{68}}\]
Hence , the required probability is \[\dfrac{3}{{68}}\]
Thus , the correct option is \[\left( C \right)\]
So, the correct answer is “Option C”.

Note: Corresponding to each combination of ${}^n{C_r}$we have $r!$permutations, because $r$ objects in every combination can be rearranged in $r!$ ways . Hence , the total number of permutations of $n$ different things taken $r$ at a time is\[{}^n{C_r} \times {\text{ }}r!\]. Thus \[\;{}^n{P_r}{\text{ }} = {\text{ }}{}^n{C_r}{\text{ }} \times {\text{ }}r!{\text{ }},{\text{ }}0 < {\text{ }}r{\text{ }} \leqslant n\]
Also , some formulas used :
\[{}^n{C_1} = {\text{ }}n\]
\[{}^n{C_2}{\text{ }} = {\text{ }}\dfrac{{n\left( {n - 1} \right)}}{2}\]
\[{}^n{C_0}{\text{ }} = {\text{ }}1\]
\[{}^n{C_n} = {\text{ }}1\]