Question

Three numbers are chosen at random without replacement from $\left\{1,2,...8\right\}$. The probability that their minimum is $3$, given that their maximum is $6$, is

• A. $\frac{3}{8}$
• B. $\frac{1}{5}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B

$\frac{1}{5}$

Explanation for the correct answer:

Let $A$ be the event that of getting minimum number as $3$

Let $B$be the event that of getting maximum number as $6$

Only $1$ and $2$ are numbers smaller than $3$. Hence the ways of selecting a set of $3$ numbers where $3$ is minimum are given as

${}^{8-3}{C}_{3-1}={}^5{C}_2$ [subtracted $3$ as we are excluding $1$ and $2$ and position of $3$ is fixed]

$\Rightarrow$ $n\left(A\right)={}^5{C}_2$

Only $7$ and $8$ are bigger numbers than $6$. Hence the ways of selecting a set of $3$ numbers where $6$ is the maximum are given as

${}^{8-3}{C}_{3-1}={}^5{C}_2$ [subtracted $3$ as we are excluding $7$ and $8$ and position of $6$ is fixed]

$\Rightarrow$ $n\left(B\right)={}^5{C}_2$

In both the cases there are ${}^8{C}_3$ ways of selecting $3$ numbers out of $8$.

$\Rightarrow$ $p\left(A\right)=\frac{{}^5{C}_2}{{}^8{C}_3}=\frac{{\displaystyle \frac{5!}{2!3!}}}{{\displaystyle \frac{8!}{3!5!}}}=\frac{10}{28}=\frac{5}{28}$

$\Rightarrow$ $p\left(B\right)=\frac{{}^5{C}_2}{{}^8{C}_3}=\frac{{\displaystyle \frac{5!}{2!3!}}}{{\displaystyle \frac{8!}{3!5!}}}=\frac{10}{28}=\frac{5}{28}$

The number of ways of selecting three numbers such that $3$ is the minimum and $6$ is maximum are ${}^2{C}_1$

So, $n\left(A\cap B\right)={}^2{C}_1$

$\Rightarrow$ $p\left(A\cap B\right)=\frac{{}^2{C}_1}{{}^8{C}_3}=\frac{2}{56}=\frac{1}{28}$

We are required to find the probability of $A$ given $B$(conditional probability)

$\Rightarrow$ $p\left(A|B\right)=\frac{p\left(A\cap B\right)}{p\left(B\right)}$

$=\frac{{\displaystyle \frac{1}{28}}}{{\displaystyle \frac{5}{28}}}$

$\Rightarrow$ $p\left(A|B\right)=\frac{1}{5}$

So, the probability of choosing three numbers from $\left\{1,2,...8\right\}$ such that $3$ is minimum given that $6$ is maximum is $\frac{1}{5}$.

Hence, option B is correct .