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)
= 48/52 x 47/51
= 2256/2652
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)
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.)
