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whatu
 2 years ago
Find the chance that the 5 people have 5 different birthdays.
Here is my workings but I think I have gone wrong somewhere
364/365*364/365*363/365*362/365*361/365= I can't seem to get it right have I calculated something wrong?
whatu
 2 years ago
Find the chance that the 5 people have 5 different birthdays. Here is my workings but I think I have gone wrong somewhere 364/365*364/365*363/365*362/365*361/365= I can't seem to get it right have I calculated something wrong?

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Jack1
 2 years ago
Best ResponseYou've already chosen the best response.0= 1  probability that 2 people have the same birthday

Jack1
 2 years ago
Best ResponseYou've already chosen the best response.0sorry, rephrase: = 1[(probability that any 2 people have the same birthday)+(probability that any 3 people have the same birthday)+(probability that any 4 people have the same birthday)+(probability that any 5 people have the same birthday)

whatu
 2 years ago
Best ResponseYou've already chosen the best response.0it is a bit like that but I am looking at five people with five different birthdays. I just want to know I am on the right track.

Jack1
 2 years ago
Best ResponseYou've already chosen the best response.0not great with probability sorry, will get: @Mertsj @amistre64 @hartnn @Callisto @mathslover these guys are the beez neez

Jack1
 2 years ago
Best ResponseYou've already chosen the best response.0maybe they're not on right now, will try: @UnkleRhaukus and @hba pretty please?

Jack1
 2 years ago
Best ResponseYou've already chosen the best response.0was thinking now that the answer is 1probability that 5 people have the same birthday...?

whatu
 2 years ago
Best ResponseYou've already chosen the best response.0ok think I have worked it i.e. 1 1 person 365 2 364 3 363 4 362 and 5 361 do we mutiple all these with 365 days excluding a leap year? that is five people with five diffrent birthdays

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0my idea is, and it could be faulty ...... \[\binom{5}{5}\left(\frac{1}{365}\right)^{5}\left(\frac{364}{365}\right)^{0}\]

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0ppppp ppppf pppfp ppfpp pfppp fpppp pppff ... i think the count for 5 people is 1+5+10+10+5+1 = 32 different scenarios

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0the probability that any given person has a birthday on a given day is 1/365, so the compliment of that is 364/365

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0...lol, never did like this particular one. satelitte has it tattoed someplace im sure

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0finding the chance that they have the same birthday might be easier; then its just 1same birthdays
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