Using combinations to find probabilities (please help!)

Question

1. A head, assistant, secretary, and treasurer are to be selected at random from 7 girls (including Carmela) and 8 boys (including Jerome). Each student can assume no more than one position. Find the probability that:

a. Carmela is chosen as head and Jerome as treasurer;

b. Carmela and Jerome are among the four officers

c. Carmela is not the head

d. Carmela is one of the four officers but not the head.

IM NOT SURE HOW TO DO THIS. ANY TIP OR AT LEAST ONE ANSWER WILL BE APPRECIATED, THANK YOU!

Answer 1

a. The prob that Carmela is chosen as head is 1/15 because there are 15 people of which any could be chosen. Then there would be a 1/14 probability that Jerome is treasurer because Carmela would have already been chosen for head. So this overall prob is (1/15)*(1/14) = 1/210.

b. This part is basically a generalization of part A. The probability of one configuration (Carmela is head, Jerome is treasurer) is 1/210. But in this case, any of the two can occupy any of the two positions. So first the (1/210) needs to be multiplied by 4C2 to reflect that they can hold any combo of 2 out of 4 positions. Then that value needs to be multiplied by 2! because for each combination (e.g head/treasurer, assistant/secretary, etc), one can be in one position and the other can be in the other. For example in the head/treasurer combination there are two configurations: Carmela is head/Jerome is treasurer and vice versa. So the final result is:

(1/210) * 4C2 * 2 = (1/210) * 6 * 2 = (12/210) = (2/35)

c. The prob that Carmela is not head is the same as (1 - prob_carmela_is_head). So this is just 1 - (1/15) = 14/15.

d. This is the same as what's the probability that Carmela is any of the officers minus the probability she is specifically the head. Out of each of the four positions, she has a 1/15 prob so therefore the prob of any of the four is 4 * 1/15 = 4/15. Then the prob she is head is just 1/15. Therefore the final result is:

4/15 - 1/15 = 3/15.

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