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by shivanigs » Sat Jul 21, 2012 12:58 am
Hi,

Request your help to understand the logic behind this question.Thanks..

5 different pairs of socks.Pick 2 singles,(P) that it's a pair.
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by theCEO » Sat Jul 21, 2012 2:17 am
shivanigs wrote:Hi,

Request your help to understand the logic behind this question.Thanks..

5 different pairs of socks.Pick 2 singles,(P) that it's a pair.
Number of ways to get the correct pair = 5 since every pair is different

Number of ways to make pairs of socks = (10 * 9)/2 = 45. (divide by 2 because order doesn't matter)

Probabilty that it is a pair = 5/45 = 1/9
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by Lifetron » Sat Jul 21, 2012 6:10 am
theCEO wrote:
shivanigs wrote:Hi,

Request your help to understand the logic behind this question.Thanks..

5 different pairs of socks.Pick 2 singles,(P) that it's a pair.
Number of ways to get the correct pair = 5 since every pair is different

Number of ways to make pairs of socks = (10 * 9)/2 = 45. (divide by 2 because order doesn't matter)

Probabilty that it is a pair = 5/45 = 1/9
I have a doubt in the total combinations that you have selected.

Lets say

Pair 1 : AA
Pair 2 : BB

If u select [1st of 1 and 1st of 2], you get AB
the same with [2nd of 1 and 1st of 2], [2nd of 1 and 2nd of 2],[1st of 1 and 2nd of 2]
With all these you get the same combination of AB. So, the total number of combinations should reduce right ?
Oly when all 10 are diff, they will form 45 diff combinations of 2.
Am I getting something wrong here !?
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by Brent@GMATPrepNow » Sat Jul 21, 2012 6:38 am
shivanigs wrote:Hi,

Request your help to understand the logic behind this question.Thanks..

5 different pairs of socks.Pick 2 singles,(P) that it's a pair.
We can also solve the question by applying probability rules.

To pick a matching pair of socks, the following must occur:
Step 1) pick any sock
Step 2) pick the next sock so that it's a match with the first sock

So, P(matching socks) = P(pick any 1st sock AND pick matching 2nd sock)
= P(pick any 1st sock) x P(pick matching 2nd sock)
= (1) x (1/9)
= 1/9

Aside: the first probability is 1, since we can pick any sock first. The second probability is 1/9, because once we choose a first sock, there are 9 socks remaining and only one of them is a match for the first sock.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Lifetron » Sat Jul 21, 2012 6:49 am
Brent@GMATPrepNow wrote:
shivanigs wrote:Hi,

Request your help to understand the logic behind this question.Thanks..

5 different pairs of socks.Pick 2 singles,(P) that it's a pair.
We can also solve the question by applying probability rules.

To pick a matching pair of socks, the following must occur:
Step 1) pick any sock
Step 2) pick the next sock so that it's a match with the first sock

So, P(matching socks) = P(pick any 1st sock AND pick matching 2nd sock)
= P(pick any 1st sock) x P(pick matching 2nd sock)
= (1) x (1/9)
= 1/9

Aside: the first probability is 1, since we can pick any sock first. The second probability is 1/9, because once we choose a first sock, there are 9 socks remaining and only one of them is a match for the first sock.

Cheers,
Brent
Hi Brent,

Can you reply to my post, the one above your's. While taking total combinations, we should consider only distinct combinations or all possible combinations !?. Can you please point out my mistake ?

Thank you ! :)
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by kallice » Fri Aug 10, 2012 2:49 am
Hi,

Also this way

Since p= success outcomes/Total outcomes

Success outcomes= 5 , coz there are 5 pairs
total outcomes of choosing two socks = 10C2

therefore P(a pair)= 5/10C2
= 5/45
= 1/9

Hope it helps.

Thanks,
K...
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