Correct Answer - C
We have,
Probability of getting a heart in a draw `=(13)/(52)=(1)/(4)`
Probability of getting a diamond in a draw `=(13)/(52)=(1)/(4)`
Probability of getting a black card in a draw `=(26)/(52)=(1)/(2)`
In 6 draws, 2 draws must contain hearts, 2 draws must contain diamond cards and 2 draws must contain black cards. This can happen in `.^6C_2xx.^4C_2xx .^C_2` mutually exculusive ways and the probability of each such way is `((1)/(4))^2xx((1)/(4))^2xx((1)/(4))^2`
Hence, required probability
`=.^6C_2xx.^4C_(2)xx((1)/(4))^2xx((1)/(4))^2xx((1)/(2))^2=90xx((1)/(2))^10`