# Probabilities in choosing committees

I got stuck on this question while tutoring a Math 12 student :

Q : There are 12 boys and 10 girls in a class. A yearbook committee of 4 is to be chosen at random. What is the probability (to the nearest 1/100th) that the committee is made up of Ryan (school president) and 3 girls?

A : The official answer says 0.2

I'm not sure if I'm calculating this correctly, my answer is :

(10C3 + 12C1) / (22C4) = (120 + 12) / 7315 = 0.0180 (which can be rounded up to 0.02)

Q : (b) if Ryan has to be on the committee, what is the probability that 3 girls are chosen?

A : The official answer says 0.09

I don't see how this is different from the previous question???

My best guess would be :

10C3 / 21C3 = 120 / 1330 = 0.0902 (which can be rounded up to 0.09)

I think the wording is what has confused me the most here, any advice/suggestions would be greatly appreciated!

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The difference between the two questions is that in the first one you have to chose the 4 persons while in the second one you only need to chose 3 because Ryan is already in the committee.

The second problem is easier. number of combinations of only girls is equal to 10C3 and the number of combinations total is 21C3 (ryan is allready in). Therefore the probability is ((10*9*8)/3!)/((21*20*19)3!) which is roughly 9%.

Ok, so the number of combinations whioch include ryan and three girls is 21*20*19 just divide this by the larger total without restrictions which is ((10*9*8)/3!)/((22*21*20*19)/4!) which is roughly 1.64%

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Thank you, Khromonkey! – Andy Jan 3 '13 at 19:43
Glad I could help. – Carry on Smiling Jan 3 '13 at 20:07

Your calculation for the first question is wrong: there are just $\binom{10}3$ ways to choose a committee consisting of Ryan and three girls, since the only freedom of choice is in the selection of the girls. The probability is therefore

$$\frac{\binom{10}3}{\binom{22}4}=\frac{120}{7315}\approx0.0164\;,$$

or $0.02$ when rounded to two decimal places. It does, therefore, appear that the official answer slipped the decimal point.

The second question asks for a conditional probability: given that Ryan is on the committee, what is the probability that the other three committee members are girls. There are $\binom{21}3$ committees that include Ryan as one of their members, and as we saw in the first problem, $\binom{10}3$ of these have three girls. The probability is therefore

$$\frac{\binom{10}3}{\binom{21}3}=\frac{120}{1330}\approx0.09\;,$$

and in this case the official answer is correct.

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Thank you Professor Scott! My original calculation came out to 0.0164 as well - I had to get creative to get to 0.018. – Andy Jan 3 '13 at 19:24
@Andy: You’re welcome. (Too creative, in this case!) – Brian M. Scott Jan 3 '13 at 19:25

The first problem is similar to a problem you already asked. There are $\binom{22}{4}$ possible committees, by assumption all equally likely.

Now we ask how many committees there are that consist of Ryan and $3$ girls. There are $\binom{10}{3}$ ways of choosing these girls.

The second problem is more subtle. It is not clear at what stage you are. In particular, it is not clear whether the machinery of conditional probability has already been introduced. We will assume that it has not.

Your analysis is correct. Given that Ryan has to be on the committee, we are choosing $3$ people from $22$.

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Thank you Mr Nicolas! – Andy Jan 3 '13 at 19:25