Birthday problem _1_9_05_

# Birthday problem _1_9_05_ - = (365) Thus, (365)(364)(363)....

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Department of Chemical Engineering University of California, Santa Barbara Ch.E. 132C. Statistical Methods in Chemical Engineering The Famous Birthday Problem In a room containing n persons, let P(n) be the probability that they all have different birthdays. For what value of n , is P(n) =0.5? Theoretical Probabilities: Number of distinct ways for people to have different birthdays: (365)(364)(363). ..(365 - + 1) Total number of ways
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Unformatted text preview: = (365) Thus, (365)(364)(363). ..(365- +1) (365) • = • • = n n n n n P(n) Results: n P(n) n P(n) 4 0.984 24 0.462 8 0.926 28 0.346 12 0.833 32 0.247 16 0.716 40 0.109 20 0.589 48 0.039 22 0.524 56 0.012 23 0.493 64 0.003 Conclusions : For n =23, P(n) ≈ 0.5. For n >32, the probability is low (<0.25) that everyone has a different birthday. For n >60, you can safely bet a handsome sum....
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## This note was uploaded on 12/29/2011 for the course CHE 132C taught by Professor Peters during the Fall '11 term at UCSB.

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