Randomly choosing an element from the set of positive divisors of 60, the probability that it will be a prime number is:

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C

Don't you think we should count both the 2s in the prime factorisation of 60? In that case the answer turns out to be 1/3, or option (B).

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C

Why isn't 1 being considered as a divisor of 60?

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C
Why isn't 1 being considered as a divisor of 60?
The formula is to get the number to prime factor and then add 1 to each power.

The number of positive factors that you, thus, get it includes 1 too.
The 12 factors of 60 are : 1,2,3,4,5,6,10,12,15,20,30 and 60.
So it includes 1 and the number itself.
Thank you for your reply. I suppose my real question is: if we are considering 1 as a divisor of 60, then there are 4 prime numbers that are divisors of 60 (1,2,3 & 5). Therefore 4/12 = 1/3 and the answer is "B."

I'm not disputing the answer, I just don't understand where my logic breaks down.

1 is not a prime number. That is why we just take 2, 3 and 5 only.

Even if someone takes 1. What is the power of 1.
1*2^2*3*5=1^(100)*2^2*3*5