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

Options

A. `(1)/(6)`
B. `.^(12)C_(2)xx(2^(6))/(12^(6))`
C. `.^(12)C_(2)xx(2^(6)-1)/(12^(6))`
D. `(341)/(12^(5))`

Answer

D

Explanation

Since anyone birthday can fall in one of 12 months. So, total number of elementary events `=12^(6)`.
Now, any two months can be chosen in `.^(12)C_(2)` ways. The six birth days can fall in these two months in `2^(6)` ways. Out of these `2^(6)` ways there are two ways when all the six birth days fall in one month. So,
Favourable number of elementary events `= .^(12)C_(2)xx(2^(6)-2)`
Hence, required probability `=(.^(12)C_(2)xx(2^(6)-2))/(12^(6)`
`=(12xx11xx(2^(5)-1))/(12^(6))=(341)/(12^(5))`