The probability that the birthdays of six different persons will fall in exactly two calendar months

Question

Smart Achievers · Accepted Answer

(3)

$n (S) = 12^{6}$

Now, any two months can be chosen in ${}^{12}C_{2}$ ways. The six birthdays can fall in these two months in $2^{6}$ ways. Out of these $2^{6}$ ways there are two ways when all the six birthdays fall in one month so favourable number of ways is ${}^{12}C_{2} \times (2^{6} - 2)$ .

Hence required probability is

$= \frac{{}^{12}C_{2} \times (2^{6} - 2)}{12^{6}}$

$= \frac{12 \times 11 \times (2^{6} - 2)}{2 \times 12^{6}}$

$= \frac{341}{12^{5}}$

Solution

Q.
The probability that the birthdays of six different persons will fall in exactly two calendar months

1 $\frac{342}{12^{5}}$

2 $\frac{341}{12^{6}}$

3 $\frac{341}{12^{5}}$

4 $\frac{342}{12^{6}}$

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Solution

Q. The probability that the birthdays of six different persons will fall in exactly two calendar months

1 342125

2 341126

3 341125