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27 Jan 2011, 06:33
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?
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27 Jan 2011, 07:59
bhandariavi wrote:
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?

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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28 Jan 2011, 04:37
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bhandariavi wrote:
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 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
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28 Jan 2011, 07:28
Thank you Bunuel and Samidh. It is crystal clear now
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28 Jan 2011, 20:34
HI Bunuel

Could you please explain this bit to me ?

P = 1 - 7/7^3. So in this case neither of approaches is right.

I'm keen to know how the numerator is 7.

Regards,
Subhash
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28 Jan 2011, 22:41
" 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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29 Jan 2011, 03:00
subhashghosh wrote:
HI Bunuel

Could you please explain this bit to me ?

P = 1 - 7/7^3. So in this case neither of approaches is right.

I'm keen to know how the numerator is 7.

Regards,
Subhash

What is the probability that all 3 men have birthday on a specific day, for example on Sunday? P=1/7*1/7*1/7=1/7^3. Now, as there are 7 days then the probability that all 3 have birthday on a same day is 7/7^3 and the probability they don't have birthday on a same day is P=1-7/7^3=48/49.

garimavyas wrote:
" 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 ?

Again: 30/49 would be the answer if the question were asking about the probability that all 3 men have birthday on different days (for example Sam-Sunday, Rick-Monday, Husam-Tuesday) BUT if 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) then the answer is P=1-7/7^3=48/49.
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29 Jan 2011, 08:23
Yeah, understood now. Thanks a lot.
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