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27 Jan 2011, 06:33
Kudos
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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 am sorry I don't have OA for this question.Bunuel, Pkit, shrouded1, icandy, IanStewart or any other members please help me!
Solutions:
I attempted it from two approaches but got two different answers
Approach 1:
7/7 * 6/7 * 5/ 7
Approach 2:
If we consider Sunday-Mon- Tue-Wed-Th-Fri- Sat as seven different seats and we can find in how many ways three people can make an arrangement on these seats.
7! / { 3! * (7-3)!} = 7! / { 3! * 4!}
Please let me know which approach is correct. Why one is wrong and another is right? or Why both of them are wrong?
I am sorry I don't have OA for this question.Bunuel, Pkit, shrouded1, icandy, IanStewart or any other members please help me!
Solutions:
I attempted it from two approaches but got two different answers
Approach 1:
7/7 * 6/7 * 5/ 7
Approach 2:
If we consider Sunday-Mon- Tue-Wed-Th-Fri- Sat as seven different seats and we can find in how many ways three people can make an arrangement on these seats.
7! / { 3! * (7-3)!} = 7! / { 3! * 4!}
Please let me know which approach is correct. Why one is wrong and another is right? or Why both of them are wrong?
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27 Jan 2011, 07:59
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bhandariavi
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.
28 Jan 2011, 04:37
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bhandariavi
I am not sure if I am correct, but let's try...
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
