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For simplicity N peoples, all born in a april (month of 30days) are collected in a room, consider the event of atleast 2 people in the room being born on the same date of month even if in different years eg.1980 and 1985 what is the smallest N so that the probability of this exceeds 0.5 is ?

Ans is 7.

My approach.is If 1 person born on a different day .then its prob is 1/365

Rest of them born on a same day (1-1/365)^N

May be it is wrong.

But i want to find smallest value of N so. Should equate 2nd equation to 0.5 ?

I've seen some related problen in this site even after that i dont understand this problem.

I got incorrect ans..by using that concept

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Hint:

Use P(at least $2$ common) = 1 - P(all different)

For all different, first person can be born on any of $30$ days,
but next one has $29$ days available and so on to be on different days.

When will $ 1 - \dfrac{30}{30}\cdot\dfrac{29}{30}\cdot \dfrac{28}{30}....$ become $>0.5$

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Assuming calculator is allowed. Probability of $N$ people having different birth-dates will be $\frac{29}{30}.\frac{28}{30}...\frac{31-N}{30}$.

You want this probability to be less than $0.5$. So keep multiplying :

$\frac{29}{30}>0.5$

$\frac{29}{30}.\frac{28}{30}=0.9022>0.5$

$\frac{29}{30}\frac{28}{30}\frac{27}{30}=0.812>0.5$

$\frac{29}{30}\frac{28}{30}\frac{27}{30}\frac{26}{30}=0.704>0.5$

$\frac{29}{30}\frac{28}{30}\frac{27}{30}\frac{26}{30}\frac{25}{30}=0.586>0.5$

$\frac{29}{30}\frac{28}{30}\frac{27}{30}\frac{26}{30}\frac{25}{30}\frac{24}{30}=0.469<0.5$

$24$ is $31-7$. So $N$ is $7$

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hint: (30^N-30Cn*n!)/30^n >0.5 the total number of functions - one one functions = favourable number of outcomes only 7 gives 0.53

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