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A playgroup is made up entirely of n pairs of siblings, including the

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A playgroup is made up entirely of n pairs of siblings, including the [#permalink]

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A playgroup is made up entirely of n pairs of siblings, including the siblings Adam and Josh. 4 members of the playgroup are chosen to represent it in a competition. What is the value of n?

(1) The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
(2) The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5
[Reveal] Spoiler: OA
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A playgroup is made up entirely of n pairs of siblings, including the [#permalink]

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New post 07 Feb 2018, 19:08
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saswata4s wrote:
A playgroup is made up entirely of n pairs of siblings, including the siblings Adam and Josh. 4 members of the playgroup are chosen to represent it in a competition. What is the value of n?

(1) The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
(2) The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5



Hi..

If you are clear with nuances of probability, you will know that D should be the answer in most cases.

But let us solve it..
(1) The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
So 2 are already selected.
2 have to be selected from 2n-2 now..
So \(\frac{(2n-2)C2}{2nC2}=\frac{2}{5}\)........ You should get n from here..
\(\frac{2(n-1)(2n-3)}{(2n)(2n-1)}=\frac{2}{5}\).....\(10n^2-25n+15=4n^2-2n.......6n^2-23n+15=0.....(6n-5)(n-3)=0\)..
so n is either 3 or -5/6.. only 3 is possible..
Sufficient

(2) The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5
So this is almost same as choosing 2 out of n
Probability:-
\(\frac{nC2}{2nC2}=\frac{n(n-1)}{2n(2n-1)}=\frac{n-1}{4n-2}=\frac{1}{5}.......5n-5=4n-2.....n=3\)
Sufficient

D
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Re: A playgroup is made up entirely of n pairs of siblings, including the [#permalink]

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New post 08 Feb 2018, 04:31
saswata4s wrote:
A playgroup is made up entirely of n pairs of siblings, including the siblings Adam and Josh. 4 members of the playgroup are chosen to represent it in a competition. What is the value of n?

(1) The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
(2) The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5


In addition to the Precise approach shown above, we can also look for different values of n that solve the problem.
This is an Alternative approach.

There are at least 4 members in the playgroup so n is at least 2.

(1) If n = 2 then Josh and Adam must be chosen and p = 1.
If n = 3 then there are 6C4 ways to choose any 4 members and 4C2 ways to choose Josh, Adam and an additional 2 of the remaining 4
Then p = 4!/2!2! : 6!/4!2! which simplifies to 12/30 = 2/5 as required. So n = 3 is possible.
Let's see what happens when we increase n.
If n = 4 then p = 6C2 : 8C4 which simplifies to 20/42, or about half.
Looks like increasing n increases p! Let's try one last one:
If n = 5 then p = 8C2 : 10C4 which simplifies to 60/90 another increase.
(1) is sufficient.

(2) Once again, if n = 2 then p = 1, impossible.
If n = 3 then we need to choose two of three pairs or 3C2 out of 6C4 total ways to choose the members.
This gives 3!/2! : 6!/4!2! = 1/5.
Trying n = 4 would give 4C2 : 8C4 which simplifes to 12/140, less than 1/5.
Since the trend is decreasing, (2) is also sufficient.

(D) is our answer.


** small note: As per chetan2u comment above, changing the number of things we are counting almost always changes the total probability.
So, if you can create an equation then it is very likely that this equation has only one solution.
This is why, once we can see that we have sufficient information to create an equation, we should feel comfortable guessing 'Sufficient'.
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Re: A playgroup is made up entirely of n pairs of siblings, including the   [#permalink] 08 Feb 2018, 04:31
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