Question

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

• A. $\frac{1}{5}$
• B. $\frac{1}{4}$
• C. $\frac{2}{5}$
• D. $\frac{3}{8}$

Answer 1

The correct option is A $\frac{1}{5}$

Let $A$ be an event such that the maximum of the three randomly chosen numbers is $6$
Let $B$ be an event such that the minimum of the three randomly chosen numbers is $3$
$\text{Probability of occurrence of event}A=P\left(A\right)=\frac{{}^5{C}_2}{{}^8{C}_3}$
$[\text{The numbers}<6\text{are}1,2,3,4,5]$
$\text{Probability of occurrence of event}B=P\left(B\right)=\frac{{}^5{C}_2}{{}^8{C}_3}$
$[\text{The numbers}>3\text{are}4,5,6,7,8]$

The probability of choosing numbers which has minimum $3$ and maximum $6$ is
$P(A\cap B)=\frac{{}^2{C}_1}{{}^8{C}_3}[\because 3<4,5<6]$
The probability that their minimum is $3$, given that their maximum is $6$ is
$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{{}^2{C}_1}{{}^5{C}_2}=\frac{2}{10}=\frac{1}{5}$