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. $\frac{2}{5}$
• 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 ${}^{\prime }{3}^{\prime }$ and maximum probabilities is ${}^{\prime }{6}^{\prime }$ then,

we get

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

So, now we have 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 ${}^{\prime }{3}^{\prime }$ 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 $=\frac{2}{10}=\frac{1}{5}\iff P=\frac{1}{5}$