Question

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

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

Answer 1

The correct option is D $\frac{51}{92}$

We have to find the probability that the divisors of $504$ chosen are even.

First let us factor $504$.

$504=2^3\times 3^2\times 7$

We know that the number of divisors of a number $n=p^a{q}^b{r}^c...$ is given by $d\left(n\right)=(a+1)(b+1)(c+1)...$

Therefore, prime factors of $504=(3+1)(2+1)(1+1)$

$=4\times 3\times 2=24$

Hence, total number of divisors is $24$.

$\Rightarrow n\left(S\right)=24C_2$

Let $A$ denote the even divisors of $504$.

We know that, for a number we will get odd divisors only if we take zero power of $2$ (since any number ($⩾1$) of $2$ will give us an even number).

Therefore, number of odd divisors of $504=(0+1)(2+1)(1+1)=1\times 3\times 2=6$.

Hence, number of even divisors of $504=\text{total number of divisors}-\text{number of odd divisors}$.

$\Rightarrow$ number of even divisors $=24-6=18$

Therefore, $n\left(A\right)=18C_2$

Hence, $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

$=\frac{{}^{18}{C}_2}{{}^{24}{C}_2}$

$=\frac{18\times 17}{24\times 23}$

$=\frac{51}{92}$

Therefore, the probability that the even divisors are chosen is $\frac{51}{92}$.