Two people are to be selected at random from a certain group that

Given: Two people are to be selected at random from a certain group that includes Claire and Max.
Asked: What is the probability that the 2 people selected will include Claire but not Max?

(1) The probability that the 2 people will be selected will be Claire and Max is \(\frac{1}{15}\)
Let total persons including Claire and Max be n.
The probability that the 2 people will be selected will be Claire and Max = \(\frac{1}{^nC_2} = \frac{2}{n(n-1)} = \frac{1}{15}\)
n(n-1) = 30; n = 6
The probability that the 2 people selected will include Claire but not Max \(= \frac{4}{^6C_2} = \frac{4}{30} = \frac{2}{15}\)
SUFFICIENT

(2) The probability that the 2 people selected will include neither Claire nor Max is \(\frac{2}{5}\)
Let total persons including Claire and Max be n.
The probability that the 2 people selected will include neither Claire nor Max \( = \frac{^{n-2}C_2}{^nC_2 }= \frac{(n-2)(n-3)}{n(n-1)} =\frac{2}{5}\)
n=6
The probability that the 2 people selected will include Claire but not Max \(= \frac{4}{^6C_2} = \frac{4}{30} = \frac{2}{15}\)
SUFFICIENT

IMO D

Bunuel

Thank you Bunuel for your quick reply!

Two people are to be selected at random from a certain group that includes Claire and Max. What is the probability that the 2 people selected will include Claire but not Max?

(1) The probability that the 2 people selected will be Claire and Max is \(\frac{1}{15}\).

This question seems tough, but we can evaluate this statement without using any math.

Notice that there will only be one case in which this statement will be true.

If, for instance, there were just 3 people in the group, the probability that the 2 people selected will be Claire and Max would be greater than \(\frac{1}{15}\).

If there were 200 people in the group, the probability that the 2 people selected will be Claire and Max would be much less than \(\frac{1}{15}\).

Only in one particular case will the probability be \(\frac{1}{15}\), and knowing that, we could work from that information to the size of the group and then to the probability that the 2 people selected will include Claire but not Max.

Sufficient.

(2) The probability that the 2 people selected will include neither Claire nor Max is \(\frac{2}{5}\).

As is the case with statement (1), this statement will be true only of the group is of one particular size.

So, we could work from this probability to the size of the group and then to the probability that the 2 people selected will include Claire but not Max.

Sufficient.

Correct answer: D­

Hi Bunuel,

Can you explain the multiply by 2 part. I don't really get the logic behind it. The order in which claire is choosen does not matter right?

It’s not about the order of the people in the pair but rather about the two different scenarios of selecting Claire and not Max:

• You first choose Claire and then someone who is not Max, or
• You first choose someone who is not Max (and not Claire), and then you choose Claire.

Consider this:

• If the question were asking for the probability of selecting Claire first and then someone who is not Max, the probability would be 1/n * (n - 2)/(n - 1).
• If the question were asking for the probability of selecting someone who is not Max (and not Claire) first, and then Claire, the probability would be (n - 2)/n * 1/(n - 1).

Since the question asks for the probability without specifying the order, it is the sum of these two probabilities:

1/n * (n - 2)/(n - 1) + (n - 2)/n * 1/(n - 1) = 2 * 1/n * (n - 2)/(n - 1).

This multiplication by 2 reflects the inclusion of both scenarios, not the order of the pair.

Hope it's clear.

We need to find the probability that 2 people selected will include Claire but not Max. All we need is the total number of members in the group to get this probability.

(1) The probability that the 2 people will be selected will be Claire and Max is \(\frac{1}{15}\)

We know that we can select Claire and Max in 1 way. So 15 must be the number of ways in which we can select 2 of the total n people.
nC2 = 15 which means n*(n-1)/2 = 15 so n(n-1) = 30. We can see that n = 6 is the only possible value.

This is sufficient alone to give us the required probability.



(2) The probability that the 2 people selected will include neither Claire nor Max is [m]\frac{2}{5}

This means (n-2)C2 / nC2 = 2/5 i.e. (n-2)/(n-3)/n(n-1) = 2/5

We may worry here that we will get a quadratic which might have 2 acceptable integer values. So will we solve this to check? No.

Think about it this way: We know that n = 6 will work here also so if 4*3/6*5 = 2/5, what happens when one more person is added?
We get 5*4/7*6. Comparing this fraction with the previous one, we see that the numerator has increased by 66% and denominator by 40%. So the fraction will increase. Similarly, as we keep adding people, the fraction will increasing. As we keep reducing the number of people, the fraction will keep reducing. Hence there will be only 1 positive integer value for n.

This is sufficient alone to give us the required probability.

Answer (D)

Video on Probability: https://youtu.be/0BCqnD2r-kY