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

Question

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 will be selected will be Claire and Max is \(\frac{1}{15}\)
(2) The probability that the 2 people selected will include neither Claire nor Max is \(\frac{2}{5}\)

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Bunuel · Accepted Answer

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?

Observe that the question essentially asks us to find the number of people in the group. If we knew that number, say it's n, then the probability would be  2*\frac{1}{n}*\frac{n-2}{n-1} , representing choosing Claire from n, and choosing any but Claire and Max from the remaining n-1 people. We multiply by 2 because we can choose Claire then any, or any then Claire.

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

Algebraically, the above implies that  2*\frac{1}{n}*\frac{1}{n-1} = \frac{1}{15} , from which we can find that n = 6. Sufficient.

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

Algebraically, the above implies that  \frac{n-2}{n}*\frac{n-3}{n-1} = \frac{2}{5} , from which we can find that n = 6. Sufficient.

Answer: D.

KarishmaB · Answer

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

ankitmba95 · Answer

Assume total case are 15.

Intersection = 1
Outside 2 cases of either claire or max = 6 (as 2/5 = 6/15)

Thus no way to find required value. Both cases insufficient to find '?' in the diagram. Hence answer should be 'E'

Kinshook · Answer

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

rebinh1982 · Answer

Bunuel

Thank you Bunuel for your quick reply!

MartyMurray · Answer

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} .

Similarly, we can see without performing any exact calculations that, 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} .

So, in general, as the group gets larger, the probability that Claire and Max will be the two people chosen gets smaller. It has to because, the more people there are in the group, the greater the probability that someone other than them will will chosen.

Similarly, as the group gets smaller, the probability that Claire and Max will be the two people chosen must get smaller.

There is no way around that dynamic.

In other words, it's impossible for the probability that Claire and Max will be chosen to be the same for different group sizes.

Thus, 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

Nitish1103 · Answer

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?

Bunuel · Answer

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.

Rickooreoisb · Answer

Can I solve with the methodology where for statement 2 :

I say

1 - P(both) = 2/5

This will give me only one n instead of looking at quadratic where I will not be sure of whether there is unique both or two values ?

Bunuel · Answer

No, that shortcut is  not valid here .

Statement 2 gives  P(neither Claire nor Max) . Writing 1 - P(both) = 2/5 assumes that “both” is the  only  other possibility, which is false. The complement of “neither” includes  three  cases: both, Claire only, and Max only.

egmat · Answer

The key is that  the number of people must be a positive integer .

Let's say there are  n  people in the group.

Statement 1:  P(Claire AND Max) = 1/15

There's only  1  way to pick both Claire and Max.
Total ways to pick 2 people from n = n(n-1)/2

So: 1 ÷   = 1/15
This gives us:  n(n-1) = 30

Now here's the thing: we need two consecutive integers that multiply to 30. 
- 5 × 6 = 30 ✓
- That's  n = 6

No other pair of consecutive integers gives 30. Try it: 4×5=20, 6×7=42. Only 5×6 works.

Statement 2:  P(neither Claire nor Max) = 2/5

This leads to the equation: 3n2 - 23n + 30 = 0
Solving: n = 6 or n = 10/6

But wait, can you have 10/6 people in a group? No!

Since n must be a whole number,  n = 6  is the only valid answer.

Once we know n = 6, we can calculate the target probability:
P(Claire but not Max) = 4/15

Answer: D

Adit_ · Answer

This problem is basically realising 2 things:
1) Having both together is just 1 case since you are also choosing only 2 people. That basically means you can find HOW the denominator arrived. 
2)Also realising that extrapolating numerator and denominator by a larger number by multiplying will pull the values apart.

Statement 1:
Like I had mentioned there is only 1 case for such a thing so denominator need not be multiplied anyway here. 
nc2 = 15 and n=6 and you can find the probabilitiy of including Claire but not Max.
SUFFICIENT.

Statement 2:
This is interesting.
2/5 is the given probability.
Now we aren't taking help of 1st statement here.
We can still say that:
n-2c2 / nc2 = 2/5.
Here you can again see that n is a unique value = 6 here.
SUFFICIENT.

Answer:  Option D

_______________________________

Like others have pointed out the direct math, it might not be easy to conclude that when you could be facing a quadratic and unsure if you wanna solve it for DS. The way you can do this is, stretch 2/5 to 6/15 and since you know 15=6c2 from statement 1 try checking if 4c2=6 which is true! Again we are only taking help of statement 1 to check for values to our benefit! We aren't combining the two here at all. Now once that is done you can check neighbouring values and notice that the values are stretched apart if you multiple by a common number.
Like 18/45 if you take 45 = 10c2 now 8c2 = 28 and not 18. Pushing this further you get numerator values being larger and larger and never tending to the actual numerator. Thus it is certain that we have only 1 value.

gullyboy09 · Answer

Thanks Marty for this. Now I don't need to solve the complex quadratic where simple logic can help.

AkshayMetha · Answer

This is the best one available.

Just wanted to check, 
1. will this logic work for similar probability questions as well? 
2. Are we expected to keep few checkpoints in mind where this logic might not work before exam? 
3. How is this working in spite of fact that (if we form equation - which happens to be quadratic) for every value of output Y (given expected probability here) we should ideally have 2 values of x (i.e. possible number of people x for required probability Y) as quadratic equation is parabola. But still we deduced that probability 2/5 will only come when n=6 as input. Luckily, the other solution here was non-integer, hence, eliminating the case that n=that non-integer (1.6666) as it could not be the number of people. But had it been an integer, then?

Just wondering how to proceed in this case? Save time on exam and move ahead or get stuck and solve equation to confirm it has only one plausible solution. Thanks in Advance  MartyMurray  As usual, appreciate you and your knowledge forever!!

AkshayMetha · Answer

KarishmaB - Wanted to ask how can we be sure in exam that there would be one solution. As you mentioned, if we move the value of input n (x) the output probability ratio in this case (y) moves upward or downward. As it should do, in most of the cases, as a quadratic equation is parabola.. so there will always lie a chance that for desired Y (probability) there can be 2 input (x) values possible.

Just wanted to check how to approach / think in such DS questions. Thanks in advance

Just wanted to check how to approach / think in such DS questions. Thanks in advance

egmat · Answer

Hi AkshayMetha,

Your worry is a good one, and you're right that  as pure algebra  a quadratic gives two roots, so in principle you could fear two valid integer values of n. But here the uniqueness isn't luck - it's built into the situation, and that's exactly the point KarishmaB and MartyMurray were making in the thread.

Why there can only ever be one n

The probability in Statement  2  isn't a free-floating quadratic. It comes from a real relationship:

P(neither) = (n- 2 )(n- 3 ) /  1 )]

As the group grows, this fraction  strictly increases  - every extra person makes it likelier that neither Claire nor Max is picked. Because it's strictly increasing,  each value of the probability is hit by exactly one n.  Two different group sizes can never produce the same P(neither). So the moment a statement pins down that probability, it pins down n - and the question is settled.

Watch it move:

- n =  5  - ( 3 )( 2 )/( 5 )( 4 ) =  6/20 = 0.30 
- n =  6  - ( 4 )( 3 )/( 6 )( 5 ) =  12/30 = 0.40 
- n =  7  - ( 5 )( 4 )/( 7 )( 6 ) =  20/42 ≈ 0.476

The values never repeat. So  2/5  can match only one of them.

What about the second root?

The "other" root (your  10/6 ) is just an  algebraic shadow  of forcing the relationship into quadratic form. The real domain is positive integers with n ≥  4 , and the monotonic structure guarantees only one root can live there - it will always fall out as non-integer, negative, or too small. It's not a coincidence you can't rely on; it's forced.

On the exam

Don't factor the quadratic. Once you recognize the probability moves in one direction with n, you can declare  sufficient  immediately and move on. That recognition - not the algebra - is what the question is testing.

Answer: D

KarishmaB · Answer

When you have a quadratic in hand, the sum and product of the roots using Vieta's formula help you gauge whether you will have only one positive solution or two are possible. That much info could be enough to know whether you have a unique value since in real life contexts, negative values are likely ignored.

e.g. n(n-1) = 30
n^2 - n - 30 = 0
Sum of roots = 1, Product = - 30
Since product is negative, one root is positive and one is negative. So n has only 1 possible positive value.

On the other hand,  \frac{(n-2)(n-3)}{n(n-1)} = \frac{2}{5}  would require some manipulation to arrive at the quadratic in the first place. This should be an indicator that we don't really need to solve it. 
So notice the pattern: 
When n = 6,  \frac{(n-2)(n-3)}{n(n-1)} = \frac{2}{5} 
When n = 7,  \frac{(n-2)(n-3)}{n(n-1)} = \frac{20}{42} 
...

As n increases, the fraction increases. Note that it makes sense since as the number of people increases, the probability that 2 specific people are not included will keep increasing. If there are 10 people, then there is a certain probability that 2 specific people are not included. But if there are 100 people, then there is a much higher probability that those 2 specific people are not included. For no value of n greater than 10 can this value be less than that obtained for n = 10.

MartyMurray · Answer

Hi  AkshayMetha

I personally would only consider the logic of the situation to confirm without solving that there is only one plausible solution and move on.

Yes it will, but we have to consider the unique logic of each situation to see whether this type of logic will work for a particular question.

I personally would not. I think it's better to have a general understanding of the logic of probability that can be applied flexibly.

That said, having a clear understanding of why this logic works in this case and does not in others would be good.

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

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