Probability of both members of a committee being girls

Question

A small committee of two is formed from a group of 4 boys and 8 girls. If at least one of the members of the committee is a girl, what is the probability that both members are girls?

So far I have tried: Multiplying the original number of original girls ($\frac{8}{12}$) by one minus hence it is given to us one is already part of the committee ($\frac{7}{11}$) to no avail. Also did some other things similar to what i just said before that I don't remember since I was trying everything that I could conjure up.How do i tackle this problem?.

Answer 1

Define event $A$ to be both members are girls and $B$ to be at least one member is a girl, we have

$$P(A) =1- \frac{{4 \choose 2}}{{12 \choose 2}} = \frac{10}{11}$$

$$P(B) = \frac{{8 \choose 2}}{{12 \choose 2}} = \frac{14}{33}$$

So, you are looking for $$P(A |B ) = \frac{P(A \cap B)}{P(B)} \stackrel{A \subset B}{=} \frac{P(A)}{P(B)} = \frac{\frac{14}{33}}{\frac{10}{11}} = \frac{7}{15}$$

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This simulation in python seems to confirm the result

from random import randint
from itertools import combinations

sample_space = list(combinations([x for x in range(1,13)],2)) # all the possible couples

r,s = 0,0 # r = numerator and s = denominator of P(A|B)
q = len(sample_space)

for _ in range(100000):

    couple = sample_space[randint(0,q-1)] # randomly selecting a couple

    if couple[0] <= 8 or couple[1] <= 8: s += 1 # if at least one person = girl increment den.
    if couple[0] <= 8 and couple[1] <= 8: r += 1 # if both person = girls increment num.

because $r/s \approx \frac{7}{15}$.

Edit :

I made the simulation because the probability seemed too low to me, then thanks to @Bram28 answer I realized that another way was possible, and that was also a better intuitive estimate of the amount I had in mind !

Answer 2

Since one girl is already on the committee there are 4 boys and 7 girls left to choose from. If the choice is random then the probability that the second person is also a girl is 7/(7+ 4)= 7/11.

Answer 3

The probabilities: $$P(GG)=\frac8{12}\cdot\frac7{11}; P(GB)=\frac8{12}\cdot\frac4{11};P(BG)=\frac4{12}\cdot\frac8{11};P(BB)=\frac4{12}\cdot\frac3{11}.$$ Hence: $$P(GG|GG\cup GB \ \cup BG)=\frac{P(GG)}{P(GG\cup GB \ \cup BG)}=\frac{\frac8{12}\cdot\frac7{11}}{1-\frac4{12}\cdot\frac3{11}}=\frac7{15}.$$

Answer 4

First off, when you multiply $\frac{8}{12}$ by $\frac{7}{11}$, you are computing the probability of getting two girls on the committee, period, i.e. without knowing anything other than that two people from this group of 12 will be randomly picked.

But in this problem you are given that there is at least one girl on the committee, and so that should make the probability of getting two girls more likely. That is, you didn't take into account all the given information, and probability is ultimately about what you know and what you don't know. E.g. if you know that both members are girls, then the probability that both are girls is $1$.

But thinking about this in terms of what you know and what you don't know will actually reveal a bit of an ambiguity in the problem statement. In particular, we can ask: How exactly is it that you know that 'at least on of the committee members is a girl'? Depending on the answer to that, the answer to this problem will be different.

Consider:

Scenario 1A: You were there when they formed a committee and the first random pick is a girl. So now you know that 'at least on of the committee members is a girl' ... and the probability that both members will be girls is now indeed $\frac{7}{11}$ as stated by @user247327

Scenario 1B: You weren't there when they picked both members of the committee, but one day you run into a girl who says she is on that committee. So, once again you know that 'at least on of the committee members is a girl' ... and the probability that both members will be girls is again $\frac{7}{11}$. (it is isomorph to scenario 1B)

Scenario 2: They randomly picked two people for the committee, but all you know is that this committee was invited to the Not All-Male Committees Convention of South East Alabama. So, once again you know that 'at least on of the committee members is a girl' ... but this time the probability that both members will be girls will be $\frac{7}{15}$. This is the answer given by @Tortar and @farruhota

So, I would say there are two different, but both reasonable, answers to this problem ...

Now .... given that this is undoubtedly a question from a textbook on probability, it is most likely that this is a question on conditional probability $P(A|B)$ with $A$ being the event of there being two girls on the committee, and $B$ the event that there is at least one girl on the committee ... and if we state these two events separately like that, it is intuitive to interpret event $B$ as the event of not getting an all-male committee, and that gets you to the second answer.

Another argument for picking the second answer is that this makes the 'minimal' amount of assumptions as to what you know: in the first scenario there is a specific person on the committee that you know to be a girl (again, because you ran into this girl, or saw this girl being picked first, or because the 'president' of the committee is a girl, or ...) but in the second all you know is that the committee is not all-male. Indeed, whenever there is an ambiguity like this, it is always safer to with with the interpretation that makes a minimal amount of assumptions, and so again that points to the second answer.

However (and to me, this is a big however, since I always look at the potential application of mathematics to real life), as the different scenarios I described above imply, in real life, you are probably more likely to be in a scenario compatible with the first answer than the second .... OK, ok, of course I exaggerated with my South East Alabama Not All-Male Committees Convention ... but really: it is not that easy to come up with a real-life scenario where all you know that the committee is not all-male and yet still completely randomly picked.

So, if I were you, I would first provide the second answer so as to make your textbook and teacher happy. But if you feel your teacher is the kind of person that is open to ambiguities, provide the first answer as well ... but make sure to really carefully explain the ambiguity in the question!