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IB DP Maths AI · Topic 4 Statistics and probability · SL 4.5–4.9 · ~6 min read

Probability, binomial and normal distributions

Probability rules, expected numbers of outcomes, Venn, tree and sample space diagrams, conditional probability and independence, discrete random variables and fair games, the binomial distribution, and the normal distribution with inverse normal calculations, all with GDC methods.

Probability rulesExpected number of outcomesVenn diagramsTree diagramsSample spacesConditional probabilityIndependenceDiscrete random variables
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No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Probability: P(A) = n(A)n(U) for equally likely outcomes; P(A′) = 1 − P(A). The expected number of times A happens in n trials is np.
  • Combined events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and P(A ∣ B) = P(A ∩ B)P(B). A and B are independent if P(A ∩ B) = P(A)P(B).
  • Discrete random variable: probabilities add to 1, and E(X) = Σ x P(X = x). A game is fair if the expected gain is 0.
  • Binomial X ~ B(n, p): n fixed independent trials, each with success probability p. E(X) = np, Var(X) = np(1 − p).
  • Normal X ~ N(μ, σ2): a symmetric bell curve. Use the GDC's normal cdf for probabilities and inverse normal for values.

By the end you can

  • Find probabilities and expected numbers of outcomes.
  • Use Venn diagrams, tree diagrams, sample space diagrams and tables of outcomes.
  • Use conditional probability and test for independence.
  • Find the expected value of a discrete random variable and decide whether a game is fair.
  • Recognise and use the binomial distribution.
  • Use the normal distribution, including inverse normal calculations.

1Probability rules

P(A ∪ B) = P(A) + P(B) − P(A ∩ B), P(A ∣ B) = P(A ∩ B)P(B) FORMULA BOOKLET
addition rule and conditional probability
P(A ∩ B) = P(A)P(B) FORMULA BOOKLET
independent events
Worked example

Of 120 students, 70 play a sport, 50 play music and 25 do both. Find P(sport or music) and P(music ∣ sport).

  1. sport only 45, music only 25, both 25, neither 25 M1
  2. P(S ∪ M) = 95120 = 0.792 A1
  3. P(M ∣ S) = 2570 = 0.357 A1
Worked example

Are "plays sport" and "plays music" independent?

  1. P(S)P(M) = 70120 × 50120 = 0.243 M1
  2. P(S ∩ M) = 25120 = 0.208 ≠ 0.243, so not independent R1
  • Expected number: if P(rain) = 0.3 each day, the expected number of rainy days in 30 is 30 × 0.3 = 9.
  • Tree diagrams: multiply along the branches, add the end results. "Without replacement" changes the second-stage probabilities.
Common mistakeCalling mutually exclusive events independent. Mutually exclusive means P(A ∩ B) = 0; independent means P(A ∩ B) = P(A)P(B). They are different ideas.

2Discrete random variables

E(X) = Σ x P(X = x) FORMULA BOOKLET
the long-run average
Worked example

A spinner game costs $1.50. You win $5 with probability 0.1, $2 with probability 0.3 and nothing otherwise. Find the expected gain and decide whether the game is fair.

  1. E(winnings) = 5(0.1) + 2(0.3) + 0(0.6) = $1.10 M1A1
  2. E(gain) = 1.10 − 1.50 = −$0.40 per game A1
  3. not fair: players lose 40 cents a game on average R1

3The binomial distribution

X ~ B(n, p): E(X) = np, Var(X) = np(1 − p) FORMULA BOOKLET
mean and variance
  • Conditions: a fixed number of trials; two outcomes; constant probability; independent trials.
Worked example

15% of the bulbs a factory makes are faulty. A box of 20 is chosen at random. Find P(exactly 3 faulty), P(at most 2 faulty), P(more than 3 faulty) and the expected number faulty.

  1. X ~ B(20, 0.15) A1
  2. P(X = 3) = 0.243 (binomial pdf) A1
  3. P(X ≤ 2) = 0.405 (binomial cdf) A1
  4. P(X > 3) = 1 − P(X ≤ 3) = 0.352; E(X) = 20 × 0.15 = 3 A1A1
Common mistakeTyping P(X > 3) as 1 − P(X ≤ 4). "More than 3" starts at 4, so it is 1 − P(X ≤ 3). Write the inequality out in words first.

4The normal distribution

4644764885005125245360.7500.0478μ = 500X ~ N(500, 12²)blue: 490 to 520red: below 480mass (g)
Shade the area you want before pressing any buttons. The total area under the curve is 1.
  • Shape facts: symmetric about μ; about 68% of values lie within one σ of μ, 95% within 2σ and 99.7% within 3σ.
Worked example

Bags of rice have mass X ~ N(500, 122) g. Find P(X < 480), P(490 < X < 520) and the mass exceeded by only 5% of bags.

  1. P(X < 480) = 0.0478 A1
  2. P(490 < X < 520) = 0.750 A1
  3. P(X < k) = 0.95 ⇒ k = 520 g (519.7) M1A1
Worked example

The masses have mean 500 g. The packer wants only 5% of bags above 520 g. Find the standard deviation needed.

  1. P(Z > 20σ) = 0.05, so 20σ = 1.645 (inverse normal for N(0, 1)) M1A1
  2. σ = 12.2 g A1
  • z-scores: z = x − μσ tells you how many standard deviations x is from the mean. It is how you compare results from different tests, and how you find an unknown μ or σ.
GDC (calculator) methodNormal cdf needs the lower bound, the upper bound, μ and σ (not σ2). Use −1099 or 1099 for an open end. Inverse normal uses the area to the left by default.
Linking questionThese distributions are the base for Chi-squared and t-tests, and for HL the central limit theorem, confidence intervals and the Poisson distribution (Estimation, Poisson, hypothesis tests and Markov chains). In Maths AA the same distributions come with continuous random variables: see Random variables, binomial and normal distributions.
Idea for your IA / explorationTest whether a real data set is normal (heights, reaction times, the mass of 50 packets of crisps) by comparing it with the 68–95–99.7 rule, or analyse whether a real game (a carnival stall or an online loot box) is fair.

Test yourself — cover the answers first

Quick check

  1. A fair die is rolled 60 times. How many sixes do you expect?
  2. P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Are A and B independent?
  3. X ~ B(10, 0.4). Find E(X) and Var(X).
  4. What is P(X ≥ 1) in terms of P(X = 0)?
  5. About what percentage of a normal distribution lies within μ ± 2σ?

Exam-style

  1. Test scores are N(μ, 82). 20% of students score more than 75. Find μ.
  2. A basketball player scores 70% of free throws. In 10 throws, find the probability she scores at least 8, and the most likely number of baskets.
Show answers
  1. 10
  2. yes, 0.4 × 0.5 = 0.2
  3. 4 and 2.4
  4. 1 − P(X = 0)
  5. 95%
  6. P(Z > z) = 0.2 ⇒ z = 0.8416
    75 − μ8 = 0.8416
    μ = 68.3
  7. X ~ B(10, 0.7): P(X ≥ 8) = 1 − P(X ≤ 7) = 0.383
    check the binomial pdf table: P(X = 7) = 0.267 is the largest, so 7
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.