Let S be the set of permutations of (1,2,3,4,5) whose first term is a prime.

If we choose a permutation at random from S, what is the probability that the third term is equal to 2?

thx

Guest May 28, 2021

#1**+2 **

The set S contains 72 numbers. Since the restriction is that the first term must be a prime, only 2, 3, and 5 work. 3 * 4 * 3 * 2 * 1 = 72

Now, if we need the third term to equal 2, we can get 2 * 3 * 1 * 2 * 1, which equals 12.

12/72 = 1/6

:>

ExpireNight May 28, 2021

#1**+2 **

Best Answer

The set S contains 72 numbers. Since the restriction is that the first term must be a prime, only 2, 3, and 5 work. 3 * 4 * 3 * 2 * 1 = 72

Now, if we need the third term to equal 2, we can get 2 * 3 * 1 * 2 * 1, which equals 12.

12/72 = 1/6

:>

ExpireNight May 28, 2021