# Chance of Selecting Twins from Group of 30 [Solved!]

• tomwilliam
In summary: The probability of not getting the twins is n!/(n-r)! = 30!/3!27! = 30*27*26/6 = 10*27*26. The probability of getting any other individual is n!/(n-r)! = 30!/3!27! = 30*27*25/6 = 10*27*25.This was really close - all you had to do was take account ofthe order shouldn't matter.by multiplying by the number of orderings of three choices.I would
tomwilliam
Thread moved from the technical math forums to the schoolwork forums
TL;DR Summary: Chance of picking 2 named people when randomly choosing 3 from a group of 30.

For my daughter's homework question:

There is a group of 12 girls and 18 boys. Two of them are twins (girl and boy). If I select three at random, what is the chance that the twins will be chosen?

I thought that it would be
P = 1/30 x 1/29 x 28/28
Because the first two choices have to be those specific named individuals, but then the third choice can be anyone. Now I'm not sure, because the order shouldn't matter.
Is this solution correct?
Thanks!

On your first pick, what are the chances that you will pick one of the twins?

scottdave and tomwilliam
Ah!
Is it
P=2/30 x 1/29 x 28/28?

As a sanity check, notice that the way you calculated it, having a third choice didn't change the probability at all. It just added a multiplier of 28/28=1. Does that seem logical? It seems like having a third choice should have made it more likely to get the twins.

scottdave, hutchphd, haushofer and 2 others
Yes, that's what makes me think I've got it wrong.
I think what I have is the same as the probability of selecting the two twins when choosing two out of the thirty. I have an extra available slot, but how do I factor that into the probability?

tomwilliam said:
Yes, that's what makes me think I've got it wrong.
I think what I have is the same as the probability of selecting the two twins when choosing two out of the thirty. I have an extra available slot, but how do I factor that into the probability?
Well, one of your three picks has to be the first twin another one of the three has to be the second twin and then the third pic can't be either twin. The problem is that the order is indeterminate. Maybe make a table showing the possibilities.

Thanks to all for your help.
So now I'm looking at all of the possible permutations of
Twin 1...Twin 2....Any of the other 28.
I figure that it could be 6 different permutations x 28 for each of the others. That would give me 162 permutations which are favourable, divided by 4060 possible outcomes.
P = 0.04

My daughter has decided that

P = (2C2 x 28C1)/30C3 = 0.007

But I don't see how the first C takes into account the size of the group. Also, if 2C2 = 1, then aren't we saying that the solution to this problem is the same as the simple probability of one of the three picks not being the twins?

There are CORRECTION ##30 \times 29 \times 28 = 24,360## (NOT ##^{30}C_3##) total ways of arranging three of 30 people where they are all considered different.
Of those, how many ways are there to have 2 twins, ##T_1, T_2## and one other, ##O_{1-28}## , in the first 3 positions?
You can "brute-force" the counting:
##T_1 T_2 O_{1-28}## is 28 ways.
##T_1 O_{1-28} T_2## is another 28 ways.
....
Continue to count and sum.
They are all equally likely, so you can divide your total count by CORRECTION: ##24,360## to get the probability. You can compare this answer to any formulas that you think should have given the answer more quickly.

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Thank you!
So I’m brute force calculating it and I see 6 possible ways, each of which have 28 different arrangements. This would be 168 in the numerator, out of 4060 possible outcomes = 0.04.
I’m not sure what formula to use to avoid the brute force counting: would it be Permutations? I thought about n!/(n-r)! but that doesn’t work for n=30 and r=3.

tomwilliam said:
Thank you!
So I’m brute force calculating it and I see 6 possible ways, each of which have 28 different arrangements. This would be 168 in the numerator, out of 4060 possible outcomes = 0.04.
I’m not sure what formula to use to avoid the brute force counting: would it be Permutations? I thought about n!/(n-r)! but that doesn’t work for n=30 and r=3.
I made a terrible mistake. There are 30*29*28 = 24,360 possible ways of picking 3 from 30 where order counts and all 30 are considered different.
I agree with your count of 6*28 = 168.
That would be 168/24360 = 0.0069

Last edited:
tomwilliam
tomwilliam said:
I thought that it would be
P = 1/30 x 1/29 x 28/28
Because the first two choices have to be those specific named individuals, but then the third choice can be anyone.
This was really close - all you had to do was take account of
tomwilliam said:
the order shouldn't matter.
by multiplying by the number of orderings of three choices.

tomwilliam and FactChecker
I would do it like this. The probability space is the combinations of three out of thirty. The size or cardinality of the space is "thirty chose three" which is n!/r!(n-r!) = 30!/3!27! = 30*29*28/6 = 5*29*28. All combinations are equally probable.

Of these combinations there are 28 that contain the twins. The probability of getting one of these is 28 divided by the number of combinations, the size of the probability space. 28/(5*29*28) = 1/(5*29) = 1/145.

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tomwilliam and FactChecker
tomwilliam said:
TL;DR Summary: Chance of picking 2 named people when randomly choosing 3 from a group of 30.

For my daughter's homework question:

There is a group of 12 girls and 18 boys. Two of them are twins (girl and boy). If I select three at random, what is the chance that the twins will be chosen?

I thought that it would be
P = 1/30 x 1/29 x 28/28
Because the first two choices have to be those specific named individuals, but then the third choice can be anyone. Now I'm not sure, because the order shouldn't matter.
Is this solution correct?
Thanks!
First, let's analyse the problem to see what we are talking about. Let's have G for the girl twin; B for the boy twin and X for anyone else. It's a success if we get both of the twins. The cases are:

GBX
GXB
BGX
BXG
XGB
XBG

Now you have calculate the probability of each of these.

PS Once you understand what you are doing, you may start to see why binomial combinations are relevant.

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tomwilliam
Thanks to everyone for helping!

tomwilliam said:
Thanks to everyone for helping!
Here's a idea/suggestion. Usually a question like this asks you to calculate only one thing. In this case, the probability of picking both twins. This might lead to only a partial understanding of these problems. When I'm learning I like to be more thorough in terms of (within reason) looking at the whole problem. And thereby seeing the whole picture. And, why not? If the aim is to learn probability theory, why just do problems in little bits and pieces?

In this case, I would have tried to see the whole picture. Here are my thought processes.

There are four outcomes:

1) You have both twins.

2) You have only the girl twin (and two others).

3) You have only the boy twin (and two others).

4) You have neither twin.

Observation: these are mutually exclusive; and represent all the possible outcomes. So, these four probabilities must add up to ##1##.

Observation: they are not all equally likely;

Intuitive insight: 2) and 3) must have an equal probability. In more complicated problems, I would think carefully about relying on these insights. So, I might calculate 2) and 3) in any case and check they are the same. But, in this case, it seems clear to me that 2) and 3) must be equal.

My next thought is what is the easiest to calculate? It looks like 4). That must be:
$$p_4 = \frac {28} {30} \times \frac {27} {29} \times \frac {26} {28} = 0.807$$And that seems like a sensible number.

Intuitive insight: the probability of getting the girl on the first choice is the same as same as getting the girl on the second choice, and the same on the third choice. This insight might not be so easy to see. Let's calculate and see:
$$p(GXX) = \frac {1} {30} \times \frac {28} {29} \times \frac {27} {28} = 0.031$$$$p(XGX) = \frac {28} {30} \times \frac {1} {29} \times \frac {27} {28} = 0.031$$$$p(XXG) = \frac {28} {30} \times \frac {27} {29} \times \frac {1} {28} = 0.031$$And we see that it's the same probability in all three sub-cases. So:
$$p_2 = p(GXX) + p(XGX) + P(XXG) = 3 \times 0.031 = 0.093$$Now, it should be clear that we have the same calculation for ##p_3##. If we do the calculution, we are simply replacing the G with a B. So:$$p_3 = 0.093$$
This now gives us a method of calculating ##p_1##, using the fact that these are mutually exclusive and represent all the possible outcomes, hence must sum to ##1##:
$$p_1 = 1 - p_2 - p_3 - p_4 = 0.007$$
Finally, it's almost always a good idea to calculate the answer two different ways and check they are the same. So, we should calculate ##p_1## directly.

Intuitive insight: all six sub-cases of getting both twins have the same probability. That is to say:
$$p(GBX) = p(GXB) = p(XGB) = p(BGX) = p(BXG) = p(XBG)$$You can check this for yourself. In any case:
$$p(GBX) = \frac{1}{30} \times \frac {1}{29} \times \frac{28}{28} = 0.00115$$$$p_1 = 6 \times p(GBX) = 0.007$$That's how I would approach the problem if I were revising probability theory. And, I can now answer any question relating to picking one or both of the twins.

tomwilliam and DrClaude
pbuk said:
This was really close - all you had to do was take account of
Interesting. The third choice should not have increased the probability at this point, and even more choices could be added with no effect at this point.
pbuk said:
by multiplying by the number of orderings of three choices.
But then the additional choices would increase the multiplier for the number of orderings of the choices.

tomwilliam
FactChecker said:
Interesting. The third choice should not have increased the probability at this point, and even more choices could be added with no effect at this point.
No. For example, if there were 1,000,000 people in addition to the twins, the probability of drawing both of them would be insignificant. If there was only one person in addition to them, the probability of drawing both of them would be 1.0

tomwilliam said:
There is a group of 12 girls and 18 boys. Two of them are twins (girl and boy).
The girl/boy differentiation is a "red herring." I.E., information extraneous to the question. This, I think, is another valuable insight.

phinds said:
No. For example, if there were 1,000,000 people in addition to the twins, the probability of drawing both of them would be insignificant. If there was only one person in addition to them, the probability of drawing both of them would be 1.0
Sorry that I was not clear. The quote of pbuk's post #11 did not show the relevant step of the calculation. This step assumed the twins were the first two draws. There is no increase due to an increased number of draws "at this point" of the calculation. Then the probabilities would be increased greatly when this probability was multiplied by the number of ways to reorder the draws.

phinds
FactChecker said:
Sorry that I was not clear.
Actually, I think you were being clear. I was just being inattentive.

My take, calculating the probability of _ not_ selecting the twins:
You can select 3 children from the 28 non-twins. This out of a total of 3 children chosen from the totality of 30 children.

WWGD said:
My take, calculating the probability of _ not_ selecting the twins:
You can select 3 children from the 28 non-twins. This out of a total of 3 children chosen from the totality of 30 children.
As long as you don't consider the case where only one of the twins is chosen.

FactChecker
PeroK said:
As long as you don't consider the case where only one of the twins is chosen.
No, I'm considering all or nothing. Neither son , or both.

WWGD said:
No, I'm considering all or nothing. Neither son , or both.
The law of the excluded middle!

Indeed. Inclusion-Exclusion becomes a hassle pretty quickly.

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