Question

Find the smallest number of people you need to choose at random so that the probability that at least two of them were both born on April 1 exceeds 1/2.

Answer 1

At least 2 out of n people born on April 1 means either all n are not born on April 1 or exactly 1 born on April 1.

So, P(X) = 1 - P(Y) - P(Z)

where P(Y) is the probability that none are born on April 1 and P(Z) is the probability that exactly 1 is born on April 1

P(Y) = 364ⁿ/365ⁿ

P(Z) = n * 364^n-1/365ⁿ

So, P(X) = 1 - P(0) - P(1)

= 1 - 364^n-1/365ⁿ (364 - n)

P(3) = 0.016

P(50) = 0.24

P(120) = 0.517

P(119) = 0.514

P(115) = 0.502

P(114) = 0.497

So, 115 would be the answer.

Comments

@Arjun sir, there is some mistake in P(X).

P(X) would be: 1 - 364^n-1/365ⁿ (364 + n) and n would come out to be 612.257. Hence answer is  613.

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Can we do like this?

 the probability that among n people there are at least two people with the same
birthday is given by -:

$1-P\left ( n \right )$

Where $P\left ( n \right )=$$365/366\, *\, 364/366\, *\, \cdots \left ( 366-n+1 \right )/366$

In our case it must be$\Rightarrow$$1-P\left ( n \right )$$> 1/2$

I know it will take lots of mathematical calculation..!!but i want to ask is it valid?

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Can we do like this?

Yes even this is fine.