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

Answer: (B)

$\frac{1}{5}$

Answer 2

Let EventR $\quad$ (Given : ${1, 2, 3,.........8})$ A : Maximum of three numbers is 6. B : Minimum of three numbers is 3 $P\left(\frac{B}{A}\right) = \frac{P\left(B\cap A\right)}{P\left(A\right)} = \frac{^{2}C_{1}}{^{5}C_{2}} = \frac{2}{10} = \frac{1}{5}$