The probabilities for the game at the start of Section 1 can be written in a table as:
| m | 1 | 2 | 4 |
| P(M = m) |
This gives the probability distribution of M as it shows how the total probability of 1 is distributed over the possible values. The probability distribution is often denoted by p(m) .
So
p(1) = P(M = 1) = , p(2) = , p(3) =
In general, P(X = x) = p(x) , and p can often be written as a formula.
Worked Examples
The discrete random variable X has probability distribution
p(x) = for x = 1, 2, 3, ..., 8
Find E(X) and V(X).
Substituting the values 1 to 8 into the probability distribution gives
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| p(x) |
(The probability distribution is a shorter way of giving all the probabilities associated with the random variable than drawing up a table, and indeed, there is no need to write one out if you do not feel it helps.)
As expected, note that the sum of all the probabilities is 1.
| E(X) | = \sum_{\ }^{\ } x p(x) |
| = 1 × \frac{1}{36} + 2 × \frac{2}{36} + 3 × \frac{3}{36} + ... + 8 × \frac{8}{36} | |
| = \frac{(1 + 4 + 9 + ... + 64)}{36} | |
| = \frac{204}{36} | |
| = 5\frac{2}{3} |
Does this seem likely?
Well, the values five to eight have greater probabilities than one to four so the expected answer should be more than 4.
| V(X) | = \left(1^2 × \frac{1}{36} + 2^2 × \frac{2}{36} + ... + 8^2 × \frac{8}{36}\right) - \left(5\frac{2}{3}\right)^2 |
| = \frac{(1 + 8 + 27 + ... + 512)}{36} - 32\frac{1}{9} | |
| = 36 - 32\frac{1}{9} | |
| = 3\frac{8}{9} |
In the following section, you will consider some special probability distributions which have wide applicability.
Exercises
