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29 Mar 2010, 11:46
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (02:17) correct
47%
(01:58)
wrong
based on 432
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If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
A. \(\frac{85}{365} * \frac{84}{364}\)
B. \(\frac{1}{365} * \frac{1}{364}\)
C. \(1 - \frac{85!}{365!}\)
D. \(1 - \frac{365!}{280!(365^{85})}\)
E. \(1 - \frac{85!}{(365^{85})}\)
A. \(\frac{85}{365} * \frac{84}{364}\)
B. \(\frac{1}{365} * \frac{1}{364}\)
C. \(1 - \frac{85!}{365!}\)
D. \(1 - \frac{365!}{280!(365^{85})}\)
E. \(1 - \frac{85!}{(365^{85})}\)
gmatprep09
The easiest way to solve this problem is to calculate opposite probability and subtract this value from 1:
The opposite probability is that all students have the birthdays on different days:
\(\frac{365}{365}*\frac{364}{365}*\frac{363}{365}*...*\frac{281}{365}=\frac{365*364*363*...*281}{365^{85}}=\frac{365!}{280!*365^{85}}\) total 85 birthdays (first student can have birthday on any day =1=365/365, the probability that the second student will have the birthday on another day is 364/365, the probability that the third student will have the birthday not on this two days is 363/365, and so on).
So, the probability that at least two students in the class have the same birthday is:
\(1-\frac{365!}{280!*365^{85}}\).
Answer: D.
11 Apr 2011, 09:31
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gmatprep09
As Bunuel already explained:
P(At least two students have same birthday) = 1 - P(At most 0 students have the same birthday)
= 1 - P(All students have different birthdays)
How to choose 85 different birthdays out of 365 days OR choose 85 different days without repetition out of 365?
It is \(P^{365}_{85}=\frac{365!}{(365-85)!}=\frac{365!}{280!}\)
Total possibilities= (365)^85 as every student can choose from 365 days.
P(All students have different birthdays) \(=\frac{365!}{280!*(365)^{85}}\)
P(At least two students have same birthday) = 1 - P(All students have different birthdays)
\(P=1-\frac{365!}{280!*(365)^{85}}\)
Ans: "D"
