Question
A card is drawn from a pack. The card is replaced and the pack is reshuffled. If this is done six times, the probability that $$2$$ hearts, $$2$$ diamonds and $$2$$ black cards are drawn is :
A.
$$90.{\left( {\frac{1}{4}} \right)^6}$$
B.
$$\frac{{45}}{2}.{\left( {\frac{3}{4}} \right)^4}$$
C.
$$\frac{{90}}{{{2^{10}}}}$$
D.
none of these
Answer :
$$\frac{{90}}{{{2^{10}}}}$$
Solution :
The probability of getting a heart in one draw $$ = \frac{{13}}{{52}} = \frac{1}{4}$$
Similarly for a diamond, the probability $$ = \frac{1}{4}$$
The probability of getting a black card in one draw $$ = \frac{{26}}{{52}} = \frac{1}{2}$$
$$\therefore $$ the required probability $$ = {}^6{C_2} \times {}^4{C_2} \times {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{2}} \right)^2},$$ because in $$6$$ draws, $$2$$ draws will be for hearts, $$2$$ for diamonds and $$2$$ for black cards, and this selection can be done in $${}^6{C_2} \times {}^4{C_2} \times {}^2{C_2}$$ ways.