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probability question....

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Re: probability question....

by logitech » Sun Nov 16, 2008 2:00 pm
willbeatthegmat wrote:Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that you both enter the same section?


OA:[spoiler]17/33[/spoiler]
You and your buddy will be in either group A or groub B

P(A) + P(B) = (40/100 x 39/99) + (60/100 x 59/100) = 5100/9900 = 17/33
LGTCH
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by cramya » Sun Nov 16, 2008 2:10 pm
Similar approach as Logitech...

40C2+60C2/100C2 = 17/33
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Re: probability question....

by rohangupta83 » Mon Nov 17, 2008 4:21 am
logitech wrote:
willbeatthegmat wrote:Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that you both enter the same section?


OA:[spoiler]17/33[/spoiler]
You and your buddy will be in either group A or groub B

P(A) + P(B) = (40/100 x 39/99) + (60/100 x 59/99) = 5100/9900 = 17/33
logitech - typo. :)
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Re: probability question....

by logitech » Mon Nov 17, 2008 8:03 am
rohangupta83 wrote:
logitech wrote:
willbeatthegmat wrote:Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that you both enter the same section?


OA:[spoiler]17/33[/spoiler]
You and your buddy will be in either group A or groub B

P(A) + P(B) = (40/100 x 39/99) + (60/100 x 59/99) = 5100/9900 = 17/33
logitech - typo. :)
Oh man!! :oops:

Good catch.
LGTCH
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