Let X denotes a random variable of number of aces

Clearly, X can take values 0, 1 or 2 because only two cards are drawn.

Total deck of cards = 52

and total no. of ACE cards in a deck of cards = 4

Now, since the draws are done without replacement, therefore, the two draws are not independent.

Therefore,

P(X = 0) = Probability of no ace being drawn

= P(non – ace and non – ace)

= P(non – ace) × P(non – ace)

P(X = 1) = Probability that 1 card is an ace

= P(ace and non – ace or non –ace and ace)

= P(ace and non – ace) + P(non – ace and ace) = P(ace) P(non – ace) + P(non – ace) P(ace)

P(X = 2) = Probability that both cards are ace

= P(ace and ace)

= P(ace) × P(ace)

We know that,

Mean (μ) = E(X) = ΣXP(X)

Also, Var(X) = E(X²) – [E(X)]²

= ΣX²P(X) – [E(X)]²

= 0.1629 – 0.0237

= 0.1392

∴Standard Deviation = √Var(X) = √0.1392 ≅ 0.373(approx.)