Discrete Probability Distributions

Question

Create the probability distribution for each random variable described.

The number of heads showing in five tosses of a coin.

Answer 1

Let's let H be the number of heads among the five coin tosses. Then H can be equal to 0, 1, 2, 3, 4 or 5.

Note that with five coin tosses, there are 2 x 2 x 2 x 2 x 2 = 2⁵ = 32 possible combinations (TTTTT, HTTTT, THTTT, TTHTT, ..., HHHHH).

Let's first consider P(H=0). If H=0, this means that all five tosses were tails, so the only combination of tosses that will give five tails is TTTTT. So P(H=0) = 1 / 32.

Similarly, let's consider P(H=5). This means that all five tosses were heads, and again, the only combination that gives us five heads is HHHHH, so P(H=5) = 1 / 32.

Now let's look at P(H=1). This means that of the five tosses, one was heads and four were tails, so the following combinations are possible: HTTTT, THTTT, TTHTT, TTTHT, TTTTH. So P(H=1) = 5 / 32.

Similarly, for P(H=4) we have all combinations with four heads and one tail, which gives us five combinations: THHHH, HTHHH, HHTHH, HHHTH, HHHHT. So P(H=4) = 5 / 32.

Now we have P(H=2) and P(H=3) to calculate. First, let's note that P(H=2) = P(H=3). In each of the above calculations, the probability of having h heads is the same as the probability of having 5-h heads, so the probability of 2 heads is the same as the probability of 5 - 2 = 3 heads.

We also know that P(H=0) + P(H=1) + P(H=2) + P(H=3) + P(H=4) + P(H=5) = 1.

So 1/32 + 5/32 + P(H=2) + P(H=3) + 5/32 + 1/32 = 1.

Then 12/32 + P(H=2) + P(H=3) = 1, so P(H=2) + P(H=3) = 20/32.

But P(H=2) = P(H=3) so 2P(H=2) = 20/32, and P(H=2) = 10/32. And, of course, P(H=3) = 10/32.

This gives us:

H P(H=h)

0 1/32

1 5/32

2 10/32 = 5/16

3 10/32 = 5/16

4 5/32

5 1/32