Consider the random experiment of tossing a fair coin three times. Let X be the random variable representing the number of heads that will come out.

Question

a. Construct the probability distribution associated to the given situation.

b. Represent the probability distribution in a) graphically.

c. Compute for the E[X] and Var[[X].

Answer 1

Since you are only flipping the coin 3 times, the easiest way to solve this problem is to write out the 8 possible outcomes: TTT, TTH, THT, ... etc.

From there you can see how many of them get heads 0, 1, 2, or 3 times and that gives you your distribution which you can then graph with 0, 1, 2, and 3 on the x-axis.

The expected value is the weighted average of the number of heads you expect to see: E[X] = 1 * (probability of getting 1 head) + 2 * (probability of getting 2 heads) + 3 * (probability of getting 3 heads).

Once you have the expected value, you can plug these values in to the standard equation for the variance Var[X].

Answer 2

First, find the state space of X, all possible values that X can take.

s_X = 0,1,2,3

Now we construct the probability mass function as a table:

x 0 1 2 3

p(x) 1/8 3(1/8) 3(1/8) 1/8

For example, p(1) = P(X = 1) = 3(1/2)³. This is because P(HTT) is (1/2)³, and we multiply by 3 because there are 3 outcomes with 1 head: {HTT, THT, TTH}

E(X) = ∑x p(x) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 18/8.

Var(X) = E(X²) - [E(X)]². This means we need to find E(X²).

E(X²) = ∑x² p(x) = 0(1/8) + 1²(3/8) + 2²(3/8) + 3²(1/8) = 3.

Var(X) = 3 - (18/8)² = 3/4.