Question

Among the divisors of $504$, two are chosen randomly, then the probability that the divisors chosen are even, is

• A. $\frac{3}{4}$
• B. $\frac{1}{24}$
• C. $\frac{1}{18}$
• D. $\frac{51}{92}$

Answer 1

Answer: (D)

$504 = 2^{3}3^{2} 7^{1}$
$\therefore $ Total number of divisors $= \left(3 + 1\right) \left(2 + 1\right) \left(1 + 1\right)$
$ = 4 \times 3 \times 2 = 24$
$\therefore n\left(S\right) = \,{}^{24}C_{2}$
Let ‘A’ denote even divisor which are $= 3 \left(2 + 1\right) \left(1 + 1\right)$
$ = 3 \times 3 \times 2 = 18$
$\therefore n\left(A\right) = \,{}^{18}C_{2}$
$\therefore P\left(A\right) = \frac{\,{}^{18}C_{2}}{\,{}^{24}C_{2}} = \frac{18}{24} \times\frac{17}{23} = \frac{51}{92}$