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# PS -probability

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Senior Manager
Joined: 08 Nov 2008
Posts: 307
Followers: 2

Kudos [?]: 86 [0], given: 7

PS -probability [#permalink]  14 Mar 2009, 01:45
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
request your detailed explanation of attached probability question .
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tt.JPG [ 36.8 KiB | Viewed 543 times ]

_________________

"CEO in making"

Senior Manager
Joined: 08 Nov 2008
Posts: 307
Followers: 2

Kudos [?]: 86 [0], given: 7

Re: PS -probability [#permalink]  14 Mar 2009, 01:46
there was some problem in uploading this question ...Anyway B is not the right choice ..
_________________

"CEO in making"

CEO
Joined: 17 Nov 2007
Posts: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 414

Kudos [?]: 2244 [0], given: 359

Re: PS -probability [#permalink]  14 Mar 2009, 03:00
Expert's post
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, 03:00
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