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PS -probability

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Senior Manager
Joined: 08 Nov 2008
Posts: 261

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14 Mar 2009, 02:45
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Difficulty:

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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request your detailed explanation of attached probability question .

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"CEO in making"

Senior Manager
Joined: 08 Nov 2008
Posts: 261

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14 Mar 2009, 02:46
there was some problem in uploading this question ...Anyway B is not the right choice ..
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"CEO in making"

CEO
Joined: 17 Nov 2007
Posts: 3486
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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14 Mar 2009, 04:00
D

One of possible solutions:

Let's find probability of the case when birthdays don't coincide:

1) the number of options for 1-st student: 365 (no restrictions).
2) the number of options for 2-nd student: 365-1 (excepting 1-st student's birthday)
...
85) the number of options for 85-th student: 365-84 (excepting all 84 students' birthdays)

$$probability = \frac{365*(365-1)...(365-84)}{365^{85}} = \frac{365!}{(365-85)!*365^{85}}$$

Answer: $$1-probability = 1 - \frac{365!}{280!*365^{85}}$$

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Re: PS -probability   [#permalink] 14 Mar 2009, 04:00
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