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Probablity

by parveen110 » Tue Sep 16, 2014 5:56 am
In a shop there were 12 different pairs of shoes. A man picked 4 shoes randomly then what is the probablity that at least 1 pair of shoes was picked?
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by GMATGuruNY » Tue Sep 16, 2014 6:33 am
Since the GMAT would use smaller numbers, let's change the problem as follows:
In a shop there are 6 different pairs of shoes. A man picks 4 shoes randomly. What is the probability that at least 1 pair of shoes is picked?
P(at least 1 pair) = 1 - P(no pairs).

P(no pairs):
The 1st shoe selected can be any of the 12 shoes.
P(2nd shoe does not match the 1st shoe) = 10/11. (Of the 11 remaining shoes, 10 do not match the 1st.)
P(3rd shoe does not match either of the first 2 shoes) = 8/10. (Of the 10 remaining shoes, 8 do not match either of the first 2 selected.)
P(4th shoe does not match any of the first 3 shoes) = 6/9. (Of the 9 remaining shoes, 6 do not match any of the first 3 selected.)
To combine these probabilities, we multiply:
10/11 * 8/10 * 6/9 = 16/33.

Thus:
P(at least 1 pair is selected) = 1 - 16/33 = 17/33.

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https://www.beatthegmat.com/probability- ... 67668.html
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by Brent@GMATPrepNow » Tue Sep 16, 2014 6:58 am
parveen110 wrote:In a shop there were 12 different pairs of shoes. A man picked 4 shoes randomly then what is the probablity that at least 1 pair of shoes was picked?
As Mitch noted, the GMAT wouldn't use such cumbersome numbers. To show why, here's a solution using your original values.

First, recognize that P(at least one matching pair) = 1 - P(no pairs)

P(no pairs) = P(select any 1st shoe AND select any non-matching shoe 2nd AND select any non-matching shoe 3rd AND select any non-matching shoe 4th)
= P(select any 1st shoe ) x P(select any non-matching shoe 2nd) x P(select any non-matching shoe 3rd) x P(select any non-matching shoe 4th)
= 1 x 22/23 x 20/22 x 19/21
= (19)(20)/(21)(23) YEESH!

So, P(at least one pair) = 1 - (19)(20)/(21)(23)
= too much of a pain to evaluate

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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