Question

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

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

Answer 1

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

Three numbers are chosen at random without replacement from $1,2,.....,8$

maximum probability $=6$.

minimum probability $=3$.

Let us consider two events $A$ and $B$ as given two probabilities.

So, we get $P(A\cap B)$

So, as per given data minimum probabilities. Should be ${}^{\prime }{3}^{\prime }$ and maximum probabilities is ${}^{\prime }{6}^{\prime }$ then, we get.

$P(A\cap B)=3,4,5,6$

So, in between we may get two probabilities which are less than ${}^{\prime }{6}^{\prime }$

So, now we have two cases.

We get ${}^n{C}_r{=}^5{C}_2$ [${}^{\prime }{5}^{\prime }$ because given there is no replacement.]

${}^5{C}_2\Rightarrow \frac{5\times 4}{2}=\frac{20}{2}=10$.

$\therefore$ total no. of classes $=10$ for minimum probabilities ${}^{\prime }{3}^{\prime }$ and maximum probabilities ${}^{\prime }{6}^{\prime }$ no. of cases $=2$

$\therefore$ the probabilities $=\frac{Totalno.ofcasesbetween{}^{\prime }{3}^{\prime }and{}^{\prime }{6}^{\prime }}{Totalno.ofcases.}$

Probabilities $=\frac{2}{10}=\frac{1}{5}\iff P=\frac{1}{5}$

$\therefore$ correct option is $C$