Question

If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?

Answer 1

Answer: The required probability is $\dfrac{10}{21}$

Step-by-step explanation:

Since we have given that

Divisors of 64= 1,2,4,8,16,32,64

Sum would be less than 32

Since it has 7 divisors in total.

If we select any two distinct divisors from first five divisors, the sum will always less than 32.

So, the probability that their sum will be less than 32 is given by

$\dfrac{^5C_2}{^7C_2}\\\\=\dfrac{10}{21}$

Hence, the required probability is $\dfrac{10}{21}$