Question

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

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

Answer 1

The correct option is D $\frac{1}{5}$.

Name events as follows

$A:$ Having minimum $3$

$B:$ Having maximum $6$

$\Rightarrow$ $A/B:$ Having minimum $3$ given that maximum is $6$

$P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}$

$P(A\cap B)=2$(Since minimum is $3$ and maximum is $6$, only $4$ or $5$ can be selected as other number)

$P\left(B\right)={{}^{5}C}_2$ (Since $6$ is maximum, rest $2$ numbers can be selected from $1,2,3,4,5$ is ${{}^{5}C}_2$ ways)

$\Rightarrow$ $\left(A/B\right)=\frac{2}{{{}^{5}C}_2}$

$=\frac{2}{10}({{}^{5}C}_2=\frac{5\times 4}{2}=10$)

$=\frac{1}{5}$