Minimum of how many people are needed to have the probability of more

chetan2u can you please explain a little bit more?

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My approach was like this:
As, question stated that at least one of the people have to have birthday on either Monday or Tuesday, I need to figure out the probability of all of them having birthday on remaining 5 days of week (other than Monday and Tuesday), then simply asses the value whether it is less than 0.5. If it is less than 0.5 then we have our answer.
First option 2 people: probability of both of them having birthday of remaining 5 days out of 7 days : \(\frac{5}{7} * \frac{5}{7}\) =\(\frac{25}{49}\) > 0.5
2nd Option 3 people: probability of all 3 of them having birthday of remaining 5 days out of 7 days : \(\frac{5}{7}*\frac{5}{7}*\frac{5}{7} = \frac{125}{363}\)< 0.5 ---> so 3 people is our answer or option [B]

Assume no of people to be x. The probability of at least one of them is born on Monday or Tuesday is 1- the probability that all of them are born on rest of the days. The probability of one person to be born on the rest of the days=5/7. The probability that all of them are born on the rest of days=(5/7)^x. Required probability= 1-(5/7)^x, which, as per statement is greater than 1/2. So, 1-(5/7)^x=1/2 or (5/7)^x< 1/2, minimum x which satisfies this equation is 3. Hence, 3 or Option B is the answer.

The reason for probability of atleast one of two person, say A and B, to be born on Mon or tuesday is..

So out of 7 days, we are taking just two days - Monday and tuesday..
Probability of A to be born on these 2 days = 2/7
Probability of B to be born on these 2 days = 2/7
But it is possible that both are born in these two days, so we have to subtract it once ..= 2/7 * 2/7 = 4/49
Probability = 2/7 + 2/7 - 4/49

Can anyone explain why you can't just add 2/7 + 2/7 to test of the probability given a party of 2? (which is >1/2).

Why must we subtract (2^2/7^2) ? I'm not following why they can't both be born on Mon/Tues..

Thank you!

By your logic, think about what happens when we have 4 people. What is the probability that at least one of them is born on either Monday or Tuesday? You would say 2/7 + 2/7 + 2/7 + 2/7 = 8/7 (which is more than 1) but that is not possible. Probability can never be more than 1.

Think of Sets here. When we say probability that one person is born on M or T is 2/7, it includes the probability that the other person is born on any day including M and T. So the 'Both' part is double counted when we do 2/7 + 2/7. So you must subtract the probability that Both are born on M or T to get
2/7 + 2/7 - 2/7*2/7 = 24/49

OR

You can instead add these three: one is born on M or T and the other is not, the other is born on M or T but one is not, both are born on M or T

So 2/7 * 5/7 + 2/7 * 5/7 + 2/7 * 2/7 = 24/49

OR

You can find the probability that neither is born on M or T by calculating 5/7 * 5/7 = 25/49
So probability that at least one is born on M or T is simply 1 - 25/49 = 24/49 (same as before but much easier to find)


Hence, this is how I would solve it:

Probability that n people are all born on other 5 days = (5/7) * (5/7) * (5/7) ... n times
When n is 2, this becomes 25/49 (greater than 1/2)
When n is 3, this becomes 125/343 (less than 1/2)

Hence the probability with 3 people that at least one is born on M or T is 1 - 'less than 1/2' which will be more than 1/2.

Answer (B)