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.
Best Answer
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.