Question

19.

a) What is the probability that two people chosen at random were born during the same month of the year?

b) What is the probability that in a group of n people chosen at random, there are at least two born in the same month of the year?

c) How many people chosen at random are needed to make the probability greater than 1/2 that there are at least two people born in the same month of the year?

Answer 1

(a). Let the first person's birth month be x. Now, the second person must born in the same month (1/12 probability) to satisfy the given condition as both their births are independent events. So, required probability is 1/12 .

(b). Birthmonth paradox :)

We have n persons and required to find if at least two of them have same birth month. So, we find the probability of none of them have same birth month and subtract this from 1. (n <= 12 as when n = 13, surely two persons will have same birthmonth as per pigeonhole principle)

P(X') = 12 * 11 * .. (12-n+1) / 12ⁿ (As the first person has 12 months to choose from, second person has 11, and so on)

= ₁₂P_n / 12ⁿ

P(X) = 1 - P(X')

(c)

P(2) = 1 - 12*11/(12*12) = 0.083

P(3) = 1 - 12 * 11 * 10 / (12 * 12 * 12) = 0.236

P(4) = 1 - 12 * 11 * 10 * 9 / (12 * 12 * 12 * 12) = 0.427

P(5) = 1 - 12 * 11 * 10 * 9 * 8 / (12 * 12 * 12 * 12 * 12) = 0.618

So, 5 is the answer.

Comments

Thanks Arjun,
Roots lies in Combinatrics of possibilities.
Got it.

 

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@arjun sir,what would have been the answer of second part if instead of atleast two,'exactly' two was asked??

please answer this one sir..