Question

Suppose we draw 5 cards at random, without replacement, from a deck
of 52 cards (such a deck includes 4 Queens). Let X denote the number
of Queens drawn. Define some indicator random variables X1,..., X5
so that X = X1+:::+X5. Then use the random variables you created
to find E(X). Also calculate the E[X] if this time 5 cards are drawn
with replacement.

Answer 1

As instructed in the question, let us define 5 random variables : X1, X2, X3, X4 and X5, such that Xi = 1, if the ith card drawn is a queen and Xi=0 otherwise.

X = X1 + X2 + X3 + X4 + X5

For any two random variables X and Y,

E(X + Y) = X + Y

irrespective of whether or not the random variables are independent or not. This result can be extended to N random variables. Thus:

E(X) = E(X1 +X2+ X3+ X4+ X5) = E(X1) + E(X2) + E(X3) + E(X4) + E(X5)

E(X1) = 4/52 = 1/13

Similarly, the expectations of each Xi (i ranges from 1 to 5) is 1/13 due to symmetry. This is easy to verify. Take the case when i=2:

E(X2) = (4/52)*(3/51) + (48/52)*(4/51) = 4/52 = 1/13

The same can easily be verified for X3, X4 and X5.

E(X) = ΣX_i = 5*(1/13)

E(X) = 5/13 = 0.3846153846

Now consider the case when the cards are drawn with replacement.

The probability of drawing a queen is the same for each draw and is given by p = 4/52 = 1/13

There are only two possible outcomes, success or failure, and the trials are carried out with replacement. So the random variable X follows binomial distribution, with n = 5 (number of draws) and probability of success, p = 1/13.

The expectation of a binomial random variable is given by:

E(X) = np

E(X) = 5*(1/13) = 5/13 = 0.3846153846

Thus it is clear that, in this particular scenario, the expectation of the random variable does not depend on the way in which the cards are drawn, that is, with or without replacement.

Hope this has been helpful. Please provide feedback and comment if I need to clarify anything in my solution.