Question

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

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

Answer 1

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

Let A be the event that maximum is 6.
B be event that minimum is 3
P(A) = $\frac{{}^5{C}_2}{{}^8{C}_3}(thenumbers<6are5)$
P(B) = $\frac{{}^5{C}_2}{{}^8{C}_3}(thenumbers>3are5)$
$P(A\cap B)=\frac{{}^2{C}_1}{{}^8{C}_3}$
Required probability is $P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{{}^2{C}_1}{{}^5{C}_2}=\frac{2}{10}=\frac{1}{5}.$