Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 28 Aug 2016, 03:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# PS -probability

Author Message
Senior Manager
Joined: 08 Nov 2008
Posts: 302
Followers: 3

Kudos [?]: 162 [0], given: 7

### Show Tags

14 Mar 2009, 02:45
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

request your detailed explanation of attached probability question .
Attachments

tt.JPG [ 36.8 KiB | Viewed 625 times ]

_________________

"CEO in making"

Senior Manager
Joined: 08 Nov 2008
Posts: 302
Followers: 3

Kudos [?]: 162 [0], given: 7

### Show Tags

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

"CEO in making"

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 501

Kudos [?]: 3066 [0], given: 360

### Show Tags

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

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Re: PS -probability   [#permalink] 14 Mar 2009, 04:00
Display posts from previous: Sort by