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