Two cards are drawn simultaneously from well shuffled pack of 52 cards, then mean and variance of number of red cards are :

1,25/51

The correct answer is Option (3) → 1,25/51

The expected value (mean) is,

$E(x)=n×\frac{k}{N}=2×\frac{26}{52}$

$=2×\frac{1}{2}=1$

The variance,

$Var(x)=n×\frac{k}{N}×\frac{N-k}{N}×\frac{N-n}{N-1}$

$=2×\frac{26}{52}×\frac{26}{52}×\frac{50}{51}$

$=\frac{25}{51}$