Question

Three numbers are chosen 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 $\frac{1}{5}$

$A\to$ Maximum is 6
$B\to$ Minimum is 3
$P\left(A\right)=\frac{{}^5{C}_2}{{}^8{C}_3}$
$P\left(B\right)=\frac{{}^5{C}_2}{{}^8{C}_3}$
$P(A\cap B)=\frac{{}^2{C}_1}{{}^8{C}_3}$
$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{{}^2{C}_1}{{}^5{C}_2}=\frac{2}{10}=\frac{1}{5}$