-> 2 cards are drawn from a pack of 52 cards.

1) What is the probability that a King and a Queen will be selected ?

2) What is the probability that the first is a King and second is a Queen ?

Can I assume that there are 4 Kings and 4 Queens in the card batch?

1)\(p=p_{1king}*p_{2queen}+p_{1queen}*p_{2king}=\frac{4}{52}*\frac{4}{51}+\frac{4}{52}*\frac{4}{51}=\frac{32}{52*51}=\frac{8}{13*51}\)

or

\(p=p_{(king) or (queen)}*p_{opposite}=\frac{8}{52}*\frac{4}{51}=\frac{32}{52*51}=\frac{8}{13*51}\)

or

\(p=\frac{C^4_1*C^4_1*P^2_2}{P^{52}_2}=\frac{4*4*2}{52*51}=\frac{8}{13*51}\)

or

\(p=\frac{C^4_1*C^4_1}{C^{52}_2}=\frac{4*4*2}{52*51}=\frac{8}{13*51}\)

or

\(p=1-\frac{P^{44}_2+C^8_1*C^3_1+C^8_1*C^{44}_1*P^2_2}{P^{52}_2}=1-\frac{44*43+8*3+8*44*2}{52*51}=1-\frac{473+6+176}{13*51}=\frac{663 -655}{13*51}=\frac{8}{13*51}\)

2) \(p=p_{1king}*p_{2queen}=\frac{4}{52}*\frac{4}{51}=\frac{16}{52*51}=\frac{4}{13*51}\)

or

\(p=\frac{C^4_1*C^4_1}{P^{52}_2}=\frac{4*4}{52*51}=\frac{4}{13*51}\)

or

\(p_2=\frac12*p_1=\frac12*\frac{8}{13*51}=\frac{4}{13*51}\) (symmetry: QK,KQ)

_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame