Probability Distributions

Every outcome with its probability

A spinner has 10 equal sectors: 2 are marked A, 3 are marked B and 5 are marked C. Every sector is equally likely, so P(A) = 2/10, P(B) = 3/10 and P(C) = 5/10.

A list of every outcome with its probability is called a probability distribution. It can be written as a table, with the outcomes in one row and their probabilities in the row below, or drawn as a chart, with one bar for each outcome.

Ten equal sectors: 2 marked A, 3 marked B and 5 marked C. Each sector is one tenth of the spinner.

The same distribution as a bar chart, read against the scale on the left: 0.2 for A, 0.3 for B and 0.5 for C.

The probabilities add up to 1

Every spin lands on exactly one of A, B and C. So the three probabilities share out the whole of 1 between them: 2/10 + 3/10 + 5/10 = 10/10 = 1.

This gives two checks for any table that claims to be a probability distribution. Every probability must be between 0 and 1, and together they must add up to exactly 1. A table whose probabilities add to 1.1, or that holds a negative probability, is not a distribution.

The whole bar is 1, cut into ten tenths. A takes 2 of them, B takes 3 and C takes 5, and together they fill the bar with none left over.

Finding a missing probability

Now there are four outcomes. A, B and C have probabilities 2/10, 3/10 and 4/10, and the probability of D is not given. The four must add up to 1, so P(D) is 1 minus the other three: 1 − 2/10 − 3/10 − 4/10 = 1 − 9/10 = 1/10.

Decimals work the same way. If three outcomes have probabilities 0.2, 0.3 and 0.4, they add up to 0.9, and the fourth is 1 − 0.9 = 0.1.

A, B and C fill 9 of the ten tenths. The one tenth left blank belongs to D, so P(D) = 1/10.

A distribution given by a rule

Sometimes the probabilities come from a rule with an unknown constant. A random variable X takes the values 1, 2, 3 and 4, with P(X = x) = kx. Then P(X = 1) = k, P(X = 2) = 2k, P(X = 3) = 3k and P(X = 4) = 4k.

They must add up to 1: k + 2k + 3k + 4k = 10k = 1, so k = 0.1. The distribution is 0.1, 0.2, 0.3 and 0.4. Once it is known, any question about X is a sum of entries: P(X ≥ 3) = 0.3 + 0.4 = 0.7.

P(X = x) = 0.1x for x = 1, 2, 3 and 4. Each line is one step taller than the one before, and the four heights add up to 1.

The usual mistakes

Taking only one probability off 1. With A, B and C at 2/10, 3/10 and 4/10, 1 − 2/10 = 8/10 forgets B and C. All three come off the total.

Giving the sum of the known probabilities as the answer. 0.2 + 0.3 + 0.4 = 0.9 is what the three given outcomes take up. The missing probability is what is left: 1 − 0.9 = 0.1.

Not checking the total. A missing probability that comes out negative, or above 1, means an error earlier in the working.

Two applications

In the first application below, the distribution of a prize is written from the sectors of a spinner and checked against a total of 1 before it is used.

In the second, one probability in a table of deliveries is unknown, and it is found by making the four probabilities add up to 1.

Worked example: A Spinner Game at a School Fair, the Fee That Makes It Fair and the Top Prize That Leaves a Profit

Question A spinner at a school fair has 8 equal sectors. One sector pays a prize of $20, two sectors pay $5 each, and the other five pay nothing. Let X be the prize won on one spin, in dollars. (a) Find E(X), and state the fee per spin that would make the game fair. (b) The stall charges $5 a spin and wants to expect a profit of $1 a spin. Keeping the two $5 sectors, what should the top prize be?

1.Write the probability distribution of X. The sectors are equally likely, so P(X = 0) = 58, P(X = 5) = 28 and P(X = 20) = 18. Check: 58 + 28 + 18 = 1.

The 8 sectors are equally likely: 5 pay nothing, 2 pay $5 and 1 pays $20.
2.Multiply each value by its probability and add: E(X) = 0 × 58 + 5 × 28 + 20 × 18 = 10 + 208 = 308 = 3.75.

Each prize is weighted by its probability: E(X) = 10 + 208 = 3.75.
3.(a) E(X) = $3.75. The game is fair when the fee equals the expected prize, so a fair fee is $3.75 a spin.

(a) The expected prize is $3.75, so a fee of $3.75 makes the game fair.
4.For a profit of $1 a spin at a fee of $5, the expected prize must be 5 − 1 = $4. Let the top prize be $x. Then E(X) = x + 2 × 58 = x + 108, so x + 108 = 4.

At a fee of $5 and a profit of $1, the expected prize must be $4: x + 108 = 4.
5.Multiply both sides by 8: x + 10 = 32, so x = 22. (b) The top prize should be $22. Check: 22 + 108 = 4, and 5 − 4 = 1 dollar of profit a spin.

(b) x + 10 = 32, so the top prize should be $22.

Answer: (a) E(X) = $3.75, so a fee of $3.75 is fair; (b) a top prize of $22

Common mistakes

Averaging the three prizes, 0 + 5 + 203 = 8.33. The three prizes are not equally likely: five of the eight sectors pay nothing, so each value must be weighted by its probability.
Treating the expected profit as a promise on every spin. One player wins $0, $5 or $22; the $1 is the average profit over many spins.

More probability distributions problems, worked step by step →

Worked example: Deliveries to a Bakery Each Morning, Where One Probability Is Missing and the Spread Is Asked For

Question The number of deliveries X that a bakery receives in one morning has P(X = 0) = 0.1, P(X = 1) = 0.3, P(X = 2) = p and P(X = 3) = 0.2, and no other values are possible. (a) Find p and E(X). (b) Find Var(X) and the standard deviation of the number of deliveries.

1.The probabilities add up to 1: 0.1 + 0.3 + p + 0.2 = 1, so p = 1 − 0.6 = 0.4.

The four probabilities add up to 1, so the missing bar is p = 0.4.
2.(a) p = 0.4, and E(X) = 0 × 0.1 + 1 × 0.3 + 2 × 0.4 + 3 × 0.2 = 0 + 0.3 + 0.8 + 0.6 = 1.7 deliveries.

(a) E(X) = 1 × 0.3 + 2 × 0.4 + 3 × 0.2 = 1.7 deliveries.
3.Find the mean of the squares, each square weighted by its own probability: E(X²) = 0² × 0.1 + 1² × 0.3 + 2² × 0.4 + 3² × 0.2 = 0 + 0.3 + 1.6 + 1.8 = 3.7.

The mean of the squares weights 0, 1, 4 and 9 by the same probabilities: E(X²) = 3.7.
4.Subtract the square of the mean: Var(X) = 3.7 − 1.7² = 3.7 − 2.89 = 0.81.

Var(X) = E(X²) − [E(X)]² = 3.7 − 2.89 = 0.81.
5.(b) Var(X) = 0.81, and the standard deviation is √0.81 = 0.9 deliveries. Check from the distances to the mean: 2.89 × 0.1 + 0.49 × 0.3 + 0.09 × 0.4 + 1.69 × 0.2 = 0.289 + 0.147 + 0.036 + 0.338 = 0.81.

(b) The variance is 0.81 and the standard deviation is √0.81 = 0.9 deliveries.

Answer: (a) p = 0.4 and E(X) = 1.7; (b) Var(X) = 0.81, standard deviation 0.9 deliveries

Common mistakes

Writing Var(X) = E(X²) − E(X) = 3.7 − 1.7 = 2. The mean must be squared before it is subtracted: 1.7² = 2.89.
Squaring the probabilities instead of the values in E(X²). The value 2 becomes 4, and it is still weighted by its own probability, 0.4 and not 0.16.

More probability distributions problems, worked step by step →