Three numbers are chosen at random without replacement from the set $A=\{x \in Z \mid 2 \leq x \leq 11\}$.…
Three numbers are chosen at random without replacement from the set $A=\{x \in Z \mid 2 \leq x \leq 11\}$. The probability that the minimum of chosen numbers is 3 and maximum is 7 is
$\frac{1}{30}$
$\frac{1}{40}$
$\frac{1}{50}$
$\frac{1}{60}$
Solution
Here, $A=\{2,3,4,5,6,7,8,9,10,11\}$
Let $E$ be the event of choosing 3 numbers and let $F$ be the event of choosing 3 numbers in which 3 is $\min$ and 7 is $\max$
$
\begin{array}{lrl}
\Rightarrow & F & =\{(3,4,7),(3,5,7),(3,6,7)\} \\
\therefore & n(F) & =3 \\
& n(E) & ={ }^{10} C_3=120 \\
\text { Required probability }=\frac{n(F)}{n(E)}=\frac{3}{120}=\frac{1}{40}
\end{array}
$