What X means
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Probability distributions
What can happen — and how likely is it?: Toss a fair coin once. X counts the heads.
| Result | X: number of heads | Probability |
|---|---|---|
| Tails | 0 | |
| Heads | 1 |
Read P(X = 1) as a question: What is the chance of getting 1 head?
Heads is 1 of the 2 equally likely results, so P(X = 1) = 1/2.
A probability distribution is a table like this: it lists every possible number X can be and its chance of happening.
| 0 | 0.20 |
| 1 | 0.35 |
| 2 | |
| 3 | 0.15 |
Find a missing probability
Find k in the table.
Step by step
- Set the total probability equal to 1.
- Add the known probabilities and subtract from 1.
Final answer
k = 0.30, which is between 0 and 1.
Find a probability from the table: X ≥ 2 means X is 2 or 3 here.
Add their chances: 0.30 + 0.15 = 0.45.
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When probabilities come from a formula
Substitute each value of x: Put each possible number into the formula.
For x = 1, replace x with 1: (1 + 1) ÷ 9 = 2/9.
| Substitute | ||
|---|---|---|
| 1 | ||
| 2 | ||
| 3 |
Check the distribution
Is this a valid probability distribution?
Step by step
- Each probability is between 0 and 1.
- Check that they add to 1.
Final answer
Yes: both probability rules hold.
If the formula contains k: Find each probability in terms of k, then set their sum equal to 1.
Find k in a formula
P(X=x)=kx for x=1,2,3. Find k.
Step by step
- Add all three chances.
- Solve.
Final answer
k = 1/6.
Exam-style question
X has the probability distribution shown in the table. k is an unknown probability.
| x | 0 | 1 | 2 |
|---|---|---|---|
| P(X = x) | 0.2 | k | 0.3 |
Try it, then check
(a) Find k. [1]
(b) Find P(X ≥ 1). [2]
Step by step
- All chances add to 1.
- X ≥ 1 means 1 or 2.
Final answer
k = 0.5; P(X ≥ 1) = 0.8.
Additional AA exam practice
IB-style question — is it valid?
A variable X has P(X = 1) = 0.1, P(X = 2) = 0.3, P(X = 3) = 0.4, P(X = 4) = 0.2.
Show that this is a valid probability distribution.
Step by step
- Each probability is between 0 and 1, and add them.
- Since they total 1, it is valid.
Final answer
All probabilities are valid and sum to 1, so it is a valid distribution.
IB-style question — find a then E(X)
X has P(X = 0) = 0.2, P(X = 1) = a, P(X = 2) = 0.3, P(X = 3) = 0.1.
Find a, then find E(X).
Step by step
- Probabilities sum to 1 — solve for a.
- E(X) = Σ x·P(x): multiply each value by its probability and add.
- Add up the terms.
Final answer
a = 0.4; E(X) = 1.3.
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AA exam practice: two unknown probabilities
One equation comes from probabilities adding to 1. A second comes from the given expected value. Solve both together, using exact values where possible.
Worked example [5 marks]
X takes values 1, 2, 3, 4 with probabilities p, 0.3, q, 0.1 respectively. Given E(X) = 2, find p and q.
Step by step
- Add the probabilities and set the total equal to 1.
- Use E(X) = sum of value × probability.
- Subtract the first equation from the second.
- Check both conditions.
Final answer
p = 0.4 and q = 0.2.
Try the next question before opening its solution.
Your turn [4 marks]
Y takes values 0, 1, 2 with probabilities a, b, 0.2 respectively. Given E(Y) = 0.9, find a and b.
Step by step
- Use total probability.
- Use the expected value. The zero outcome contributes zero.
- Substitute back.
Final answer
a = 0.3 and b = 0.5.
Exam tip: The probabilities sum to 1; the products xP(X = x) sum to E(X). These are different equations.
Additional AA exam practice
IB-style question — mode and a conditional
A discrete random variable X has P(X = 0) = 0.1, P(X = 1) = 0.2, P(X = 2) = 0.5, P(X = 3) = 0.2.
(a) Write down the mode.
(b) Find P(X ≥ 2 | X ≥ 1).
Step by step
- (a) The mode is the value with the highest probability.
- (b) Conditional probability — restrict to X ≥ 1 and take the X ≥ 2 part of it.
Final answer
(a) mode = 2. (b) 7⁄9 ≈ 0.778.
AA exam practice: ranges of probabilities and E(X)
A parameter must make every probability valid. First find its allowed interval, then use that interval to bound the expected value. Endpoints are included when zero probabilities are allowed.
Worked example [6 marks]
X takes values 1, 2, 3, 4 with probabilities r, 0.2, q, 0.1.
(a) Express q in terms of r and find the possible values of r. [3]
(b) Find the possible values of E(X). [3]
Step by step
- (a) Both r and q must be non-negative.
- (b) Write the mean in terms of r.
- The mean decreases as r increases. Evaluate both endpoints.
- Put the smaller endpoint first.
Final answer
q = 0.7 − r; 0 ≤ r ≤ 0.7; 1.5 ≤ E(X) ≤ 2.9.
Try the next question before opening its solution.
Your turn [5 marks]
Y takes values 0, 2, 4 with probabilities p, 0.25, 0.75 − p. Find the range of p and hence the range of E(Y).
Step by step
- Make p and 0.75 − p non-negative.
- Calculate the expected value in terms of p.
- Use p = 0.75 for the minimum and p = 0 for the maximum.
Final answer
0 ≤ p ≤ 0.75; 0.5 ≤ E(Y) ≤ 3.5.
Exam tip: Check every probability. Do not assume the expected value increases with the parameter; inspect its coefficient.
Additional AA exam practice
IB-style question — find k
A distribution has P(X = x) = kx for x = 1, 2, 3, 4.
Find k, then P(X = 3).
Step by step
- Sum the probabilities and set equal to 1.
- Solve, then substitute.
Final answer
k = 0.1; P(X = 3) = 0.3.