Two boys and 3 girls were selected for the finals of a dance contést. One of the three different prizes marked A. B and C will be randomly awarded to the 5 contestants whose name is selected to win. Find the probability that the boy will win the prize marked B?

2 boys and 3 girls were selected for the finals.

There are 3 prizes, A, B, and C (assigned in order, I suppose) assigned randomly to the five contestants.

We want to find the probability that a boy wins the prize B.

So here we have two cases:

A girl wins the first prize, A, and a boy the prize B.

The probability that a girl wins the prize A is equal to the quotient between the number of girls and the total number of contestants, this is:

p = 3/5

Next, the probability that a boy wins the prize B, is equal to the quotient between the number of boys and the total number of contestants (that now is 4, because a girl already won prize A).

q = 2/4

The joint probability is:

P = p*q = (3/5)*(2/4) = 6/20 = 3/10

Now we have the other case, where a boy wins the prize A and the other boy wins prize B.

The probability that a boy wins prize A is equal to the quotient between the number of boys and the total number of contestants, this is:

p = 2/5

The probability that the other boy wins prize B is equal to the quotient between the remaining number of boys (1) and the total number of contestants (now 4), this is:

q = 1/4

The joint probability is:

P' = p*q = (2/5)*(1/4) = 2/20 = 1/10

The total probability is the sum of the probabilities for the two cases, this is:

Probability = 1/10 + 3/10 = 4/10

(note that because the prize C is assigned after prize B, its existence does not affect the probability for the previous event, as expected)

the solution is 12x-6y