Probability: practice problems with step-by-step solutions

Free Probability practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: equally likely outcomes, counting and complements, binomial distribution.

Free practice problemEasy
sum ≥ 4die 1die 2112345672234567833456789445678910556789101166789101112P = ?

Two fair dice are rolled. What is the probability P that the sum is at least 4?

2 dice≥ 4

Give the probability as a decimal between 0 and 1.

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What you practise

  1. Easyequally likely outcomes
  2. Mediumcounting and complements
  3. Hardbinomial distribution

Key formulas

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Learn Probability from scratch

1What probability is

A probability is a number between 0 and 1 that says how likely something is. 0 means it never happens, 1 means it always happens.

When all outcomes are equally likely, count them:

P(A) = number of favourable outcomes / total number of outcomes

A bag holds 3 red, 5 blue and 2 green balls, 10 in total. If you draw one at random, P(blue) = 5/10 = 0.5.

A fair die has 6 faces, so P(six) = 1/6 ≈ 0.167 and P(even) = 3/6 = 0.5.

A probability can also be written as a percentage: 0.5 is 50%. Exam answers are usually decimals between 0 and 1.

Self-check A bag holds 2 red, 8 blue and 4 green balls. You draw one at random. What is P(red)?
  1. 2/12 ≈ 0.167
  2. 2/14 ≈ 0.143
  3. 2/8 = 0.25
  4. 2

Answer: 2/14 ≈ 0.143. There are 2 + 8 + 4 = 14 balls, and 2 are red: 2/14 ≈ 0.143. Dividing by 12 counts only the other balls, but the total must include the red ones too.

2Rules for combining events

  • Complement: P(not A) = 1 − P(A). Use it for “at least one” questions: P(at least one) = 1 − P(none).
  • And, independent events: multiply. Two fair coins give P(two heads) = 0.5 · 0.5 = 0.25.
  • Or: P(A or B) = P(A) + P(B) − P(A and B). Subtract the overlap so you do not count it twice.

Drawing without replacement changes the bag. After you take out one red ball, there is one red ball fewer and one ball fewer in total.

With 5 red and 4 blue balls, P(two red) = 5/9 · 4/8 = 20/72 ≈ 0.278.

To count groups where the order does not matter, use combinations. The number of ways to choose k items from n is:

C(n, k) = n! / (k! · (n − k)!)

Here n! = n · (n − 1) · … · 1. So C(6, 4) = 720 / (24 · 2) = 15 committees of 4 people from 6.

Self-check A bag holds 5 red and 4 blue balls. You draw two without putting the first one back. What is P(both red)?
  1. 0.139
  2. 0.278
  3. 0.309
  4. 0.556

Answer: 0.278. 5/9 · 4/8 = 20/72 ≈ 0.278. Using 5/9 · 5/9 ≈ 0.309 forgets that the first red ball is no longer in the bag.

3The binomial distribution

Some questions repeat the same trial n times: test 5 components, guess 8 answers, flip a coin 3 times. Each trial is a success with probability p, independently of the others.

The number of successes X then follows a binomial distribution:

P(X = k) = C(n, k) · pᵏ · (1 − p)ⁿ⁻ᵏ
  • pᵏ: the k successes.
  • (1 − p)ⁿ⁻ᵏ: the n − k failures.
  • C(n, k): the number of different orders in which the successes can fall.

For “at most one”, add P(X = 0) and P(X = 1). For “at least one”, use 1 − P(X = 0).

Self-check You flip a fair coin 3 times. What is the probability of exactly 2 heads?
  1. 0.125
  2. 0.25
  3. 0.375
  4. 0.667

Answer: 0.375. C(3, 2) · 0.5² · 0.5 = 3 · 0.125 = 0.375. Forgetting C(3, 2) gives 0.125: that is one order only, such as HHT, but THH and HTH count as well.

4The method step by step

  1. Write down what one outcome is and how many outcomes there are in total.
  2. Translate the words: “and” means multiply, “or” means add (minus the overlap), “not” means 1 − P.
  3. For “at least one”, calculate the complement “none” first.
  4. Without replacement: lower the counts after every draw. Or count with combinations.
  5. Repeated independent trials with a fixed p: use the binomial formula.
  6. Check that the answer lies between 0 and 1.

A tree diagram helps when there are several steps: multiply along a branch, and add the branches you need.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A bag holds 4 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that at least one of them is red?

1

Use the complement

“At least one red” has three cases: red–red, red–blue and blue–red. The opposite, “no red”, has only one case: blue–blue.

P(at least one red) = 1 − P(both blue)
2

First draw

There are 6 blue balls out of 10.

P(first blue) = 6/10
3

Second draw

One blue ball is gone, so 5 blue balls are left out of 9.

P(both blue) = 6/10 · 5/9 = 30/90 = 1/3
4/10R3/9R12/906/9B24/906/10B4/9R24/905/9B30/90

In the tree, R is a red ball and B a blue one. The bold branch is blue–blue.

4

Take the complement

P = 1 − 1/3 = 2/3 ≈ 0.667
5

Check the answer

Count with combinations: there are C(10, 2) = 45 pairs, and C(6, 2) = 15 of them are blue–blue. 15/45 = 1/3, the same as before.

The three other branches of the tree add up to 12/90 + 24/90 + 24/90 = 60/90 = 2/3 as well.

Answer

P = 2/3 ≈ 0.667

6Common mistakes

  • Adding probabilities for “at least one” instead of using 1 − P(none).
  • Forgetting to lower the counts when drawing without replacement.
  • Forgetting the factor C(n, k) in the binomial formula.
  • Dividing by the wrong total, for example only the other colours.
  • Giving a probability above 1 or below 0. That is always a sign of an error.
Self-check You roll a fair die twice. What is the probability of at least one six?
  1. 1/3 ≈ 0.333
  2. 1/36 ≈ 0.028
  3. 11/36 ≈ 0.306
  4. 25/36 ≈ 0.694

Answer: 11/36 ≈ 0.306. 1 − P(no six) = 1 − (5/6)² = 1 − 25/36 = 11/36. Adding 1/6 + 1/6 gives 1/3, which counts the case of two sixes twice.

7How exam questions look

  • Easy: equally likely outcomes. One draw from a bag of coloured balls, or one roll of a die.
  • Medium: counting and complements. Committees with C(n, k), or two draws without replacement.
  • Hard: the binomial distribution. Exactly k successes, or “at most one” as P(X = 0) + P(X = 1).

Give the answer as a decimal between 0 and 1, rounded to three or four decimals.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
xp(x)1?20.2530.1540.250.15

The table gives the probability distribution of X, but one probability is missing. Find P(X = 1).

x: 1, 2, 3, 4, 5p(x): ?, 0.25, 0.15, 0.2, 0.15

Find P(X = 1)

Show hints
  1. All probabilities of a distribution add up to 1.
  2. Add the known probabilities.
  3. P = 1 − 0.75.
Show the answer and the step-by-step solution

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Sum of the known probabilities

0.25 + 0.15 + 0.2 + 0.15 = 0.75

Missing probability

P(X = 1) = 1 − 0.75 = 0.25

Answer

P(X = 1) = 0.25
Example 2Medium
xp(x)10.1520.3530.0540.350.15μ = 2.95

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4, 5p(x): 0.15, 0.35, 0.05, 0.3, 0.15

Find Var(X)

Show hints
  1. μ = Σx·p(x) = 2.95.
  2. Then E(X²) = Σx²·p(x).
  3. Var(X) = E(X²) − μ².
Show the answer and the step-by-step solution

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 1·0.15 + 2·0.35 + 3·0.05 + 4·0.3 + 5·0.15 = 2.95

Mean of the squares

E(X²) = 1·0.15 + 4·0.35 + 9·0.05 + 16·0.3 + 25·0.15 = 10.55

Variance

Var(X) = 10.55 − 2.95² = 1.848

Answer

Var(X) = 1.848
Example 3Hard
n = 12, p = 0.3: P(X = 5) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 5 correct answers?

n = 12p = 0.3k = 5

Find P

Show hints
  1. This is a binomial distribution: fixed number of independent trials, same success probability.
  2. P(X = k) = C(n, k)·pk·(1 − p)n−k.
  3. C(12, 5) = 792.
Show the answer and the step-by-step solution

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Binomial formula

P(X = 5) = 792·0.35·0.77 = 792·0.00243·0.08235 = 0.1585

Answer

P = 0.1585

Questions and answers

What is Probability?

A probability is a number between 0 and 1 that says how likely something is. 0 means it never happens, 1 means it always happens.

When all outcomes are equally likely, count them:

How do I solve Probability problems step by step?

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

What are common mistakes in Probability?
  • That is the sum of the known probabilities. The missing one is 1 minus that sum.
  • The values are not equally likely: use the fact that all probabilities add up to 1.
  • E(X²) is not the variance: subtract μ².
What do the Probability exercises cover?
  • Easy: equally likely outcomes
  • Medium: counting and complements
  • Hard: binomial distribution

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Probability worksheet with answers
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