The question is quite ambiguous.

There are three people Sam, Rick and Husam. What is the probability that they don't have birthday on a same day (Sunday, Monday, ..., Saturday)?

I think that the question simply asks the probability of an event that not all 3 men have their birthdays on a same day (meaning that all 3 have birthday on Sunday is not OK, but if 2 of them have their birthday on Sunday and the third one on Monday is OK):

P=1-\frac{7}{7^3}=\frac{48}{49}. So in this case neither of approaches is right.

But if the question were asking about the probability that all 3 men have their birthdays on different days (for example Sam-Sunday, Rick-Monday, Husam-Tuesday), then:

First one can have his birthday on any day - 7/7, the probability that the second guy won't have his birthday on the same day will be 6/7 and the probability that the third guy won't have his birthday on these two days will be 5/7, so P=\frac{7}{7}*\frac{6}{7}*\frac{5}{7}=\frac{30}{49}.

Or if you want to solve with combination then: probability=# of favorable outcomes/total # of outcomes --> total # of outcomes is 7^3 as each has 7 options for a birthday. Now, # of favorable outcomes will be: C^3_7*3!, where C^3_7 is ways to choose 3 different days out of 7 and 3! is # of different ways to assign these days to 3 people (the nominator also could be represented as P^3_7). So, P=\frac{C^3_7*3!}{7^3}=\frac{30}{49}.

So in this case your first approach would be correct and second would not. By the way second approach doesn't give a probability at all (probability must be between 0 and 1 inclusive), and it gives # of selections of 3 different items (days in our case) out of 7.

Hope it's clear.
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Thank you Bunuel and Samidh. It is crystal clear now
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" Suppose, all their birthdays are on a Sunday...

therefore, probability is = 1/7 * 1/7 * 1/7

For 7 such days, it is = 7 * 1/(7*7*7) = 1/49

Now, assuming only Sam and Rick's birthday is on a Sunday...

Therefore, probability is = 1/7 *1/7 * 6/7

For 7 days, it is = 7* 1/49 *6/7 = 6/49.

There can be 3 such combo ( S & R, R & H, S & H).

Therefore, net = 6/49 * 3 = 18/49

Therefore, total probability of same birthdays = 1/49 + 18/49 = 19/49

Therefore, required answer is = 1 - 19/49 = 30 /49 "


i think the approach is ok. that includes both the possiblity of all 3 having b'day on same day and only 2 having b'day on same day.

whats the OA ?
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Yeah, understood now. Thanks a lot.
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