Question

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals
A.\[\left( 1 \right)\]$\dfrac{1}{2}$
B.\[\left( 2 \right)\]$\dfrac{7}{{15}}$
C.\[\left( 3 \right)\]$\dfrac{2}{{15}}$
D.\[\left( 4 \right)\]$\dfrac{1}{3}$

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 balls \[ = {\text{ }}7{\text{ }} + {\text{ }}3\]
\[ = {\text{ }}10\]
Total ways of arrangements of the balls in a row \[ = {\text{ }}{}^{10}{C_3}\]
Using 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 in a row \[ = {\text{ }}\dfrac{{10{\text{ }}!}}{{\left( {{\text{ }}3!{\text{ }} \times {\text{ }}7!{\text{ }}} \right)}}\]
\[ = {\text{ }}120\]
As given in the question no two black balls are placed adjacently , then the arrangement is as shown
\[\_{\text{ }}W{\text{ }}\_{\text{ }}W{\text{ }}\_{\text{ }}W{\text{ }}\_{\text{ }}W{\text{ }}\_{\text{ }}W{\text{ }}\_{\text{ }}W{\text{ }}\_{\text{ }}W{\text{ }}\_\]
There are $6$ blank spaces between the seven white balls .
So , As given no two black balls should be adjacently placed
The total positions where the white balls can be placed are $8$.
So , the favourable outcomes are \[ = {\text{ }}{}^8{C_3}\]
Again applying the formula of combination , we get
Total favourable outcomes \[ = {\text{ }}\dfrac{{8{\text{ }}!}}{{\left( {{\text{ }}3{\text{ }}!{\text{ }} \times {\text{ }}5{\text{ }}!{\text{ }}} \right)}}\]
\[ = {\text{ }}56\]
\[The{\text{ }}required{\text{ }}probability{\text{ }} = \dfrac{{total{\text{ }}favourable{\text{ }}outcomes}}{{total{\text{ }}possible{\text{ }}outcomes}}\]
Required probability \[ = {\text{ }}\dfrac{{56}}{{120}}\]
Cancelling the terms , we get
Required probability \[ = {\text{ }}\dfrac{7}{{15}}\]
Hence , the required probability is $\dfrac{7}{{15}}$

So, the correct answer is “Option B”.

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 :
\[{}^n{C_1} = {\text{ }}n\]
\[{}^n{C_2}{\text{ }} = {\text{ }}\dfrac{{n\left( {n - 1} \right)}}{2}\]
\[{}^n{C_0}{\text{ }} = {\text{ }}1\]
\[{}^n{C_n} = {\text{ }}1\]