Question

You draw cards from a standard deck of 52 playing cards. There are 12 “face cards” in the deck (J, Q, or K). Let X be the number of drawings (with replacement) it takes until you get your first face card. Let Y be the number of drawings (with replacement) it takes until you get your fifth face card. Let Z be the number of face cards removed if you draw 10 cards without replacement.
(a) Calculate P(X = 5).

(b) Calculate E[Y ]. (

c) Calculate V ar(Z).

Answer 1

a) The probability here is computed as:

P(X = 5)

= Probability that the first 4 cards drawn are not face cards * Probability that the 5th card is a face card

= [(52 - 12)/52]4 * (12/52)

= (404*12) / 525

= 0.0808

Therefore 0.0808 is the required probability here.

b) The expected value of cards drawn such that we get 5 face cards is computed here as:

= 5/p

where p is the probability of getting a face card

Therefore 21.6667 is the expected number of cards drawn here to get 5 face cards here.

c) We are given here the random variable Z as the number of face cards removed if you draw 10 cards without replacement. As we are drawing cards without replacement here, this is a case of hypergeometric distribution with parameters given as:

Population size, N = 52,
Population successes, K = 12,
Sample size, n = 10

The variance of a hypergeometric distribution is computed here as:

Therefore 1.4619 is the required variance here.