Discrete Probability Distributions

Question

Create the probability distribution for each random variable described.

The numbers of tails showing when flipping four coins.

Answer 1

Let's let T be the number of tails among the four coin flips, so T can take on values of 0, 1, 2, 3 or 4.

Before we begin, note that with four coin tosses, there are a total of 2 x 2 x 2 x 2 = 2⁴ = 16 possible outcomes: HHHH, HHHT, HHTH, HTHH, THHH, ..., TTTT

Let's first consider P(T=0). If there were 0 tails among the four tosses, that means that all four tosses were heads (HHHH). There's only one way for there to be zero tails, so P(T=0) = 1 / 16.

Next let's look at P(T=4). Similar to P(T=0), the only way that T can equal 4 is if all four coin tosses are tails (TTTT). Again there's only one way for this to happen, so P(T=4) = 1 / 16.

Now let's look at P(T=1). This happens when there is one tail and three heads, so there are four possible combinations: THHH, HTHH, HHTH, HHHT. So P(T=1) = 4 / 16 = 1 / 4.

Next let's look at P(T=3). This happens when there are three tails and one head, so similar to T=1, we have four possible combinations: HTTT, THTT, TTHT, TTTH. So P(T=3) = 4 / 16 = 1 / 4.

Since we know that T has to be 0, 1, 2, 3 or 4, we know that P(T=0) + P(T=1) + P(T=2) + P(T=3) + P(T=4) = 1. Plugging in the probabilities we already solved for, we have 1/16 + 1/4 + P(T=2) + 1/4 + 1/16 = 1, so P(T=2) + 10/16 = 1, and P(T=2) = 6 / 16 = 3 / 8.

This gives us:

T P(T=t)

0 1/16

1 1/4

2 3/8

3 1/4

4 1/16