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Find the probability that the birth days of 6 different persons will fall in exactly two calendar months.

Ans.is 341/(12^6)

Here each person has 12 option So there are 6 persons .total no. Of ways 12^6

And out of 12 months 2 are randomly selected ..so $12$C$2$

B'day of 6 persons fall in 2 months in 2^6 ways.

Therefore requird probability is ($12$C$2$ × 2^$6$)/12^$6$

But not geting appropriate ans.

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Arrange the people in order of height. Ask the shortest person what month she was born in. Whatever she says, there are $12^5$ ways the string of birthmonths, from next shortest to tallest, can be completed.

We now count the "favourables." There are $11$ choices for the other month. And for each choice of other month, there are $2^5-1$ ways the string of birthmonths can be completed to yield a favourable, for not everybody can choose the same month as the shortest person. This gives a total of $(11)(31)=341$ favourables.

Thus the required probability is $\frac{341}{12^5}$. Note that this is different in the denominator from the answer mentioned in the post.

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Suppose you are also there:

  • $12\times11$ choices for your month and that of the other group

  • $\binom50 + \binom52 + .... + \binom54 = 31$ choices for your "month-mates"

Thus ans $= \dfrac{12\cdot11\cdot31}{12^6} = \dfrac{341}{12^5}$

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Your answer's not matching because you have not considered an extra case....there is a possibility that both person have their bday in the same calendar month... Solution.... Anyone's bday can fall in any of the months...hence sample space is 12^6... Now selecting any 2 months from 12 months in $12C2$ ways out of which there is a possibility that all of them have their bday in a single month...hence favourable outcome=$12C2*(2^6-2)$... Thereby computing it results in a value of $341/12^5$

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