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A Level Maths 9709 · Paper 5 Probability & Statistics 1 · 5.4 · ~4 min read

Discrete random variables

Probability distribution tables, expectation and variance of a discrete random variable, and the binomial and geometric distributions: when to use them, their probabilities, means and variances.

Probability distributionsΣp = 1E(X)Var(X)Binomial B(n, p)Conditions for binomialGeometric Geo(p)Mean & variance formulae
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A probability distribution lists every value with its probability; the probabilities add up to 1.
  • E(X) = Σx p and Var(X) = Σx² p − [E(X)]².
  • Binomial B(n, p): a fixed number n of independent trials, each a success with the same probability p. X counts the successes.
  • Geometric Geo(p): independent trials with constant p, until the first success. X is the trial on which it happens.

By the end you can

  • Draw up a probability distribution table and use Σp = 1.
  • Calculate E(X) and Var(X) for a discrete random variable.
  • Recognise and use the binomial distribution, including its mean np and variance np(1 − p).
  • Recognise and use the geometric distribution, including P(X > r) and its mean 1/p.

1Probability distributions

Expectation and variance

E(X) = Σx P(X = x) ; Var(X) = Σx² P(X = x) − [E(X)]² FORMULA LIST (MF19)
the standard deviation is √Var(X)
Worked example

P(X = x) = kx for x = 1, 2, 3, 4. Find k, E(X) and Var(X).

  1. k(1 + 2 + 3 + 4) = 1 ⇒ k = 1/10 B1
  2. E(X) = (1 + 4 + 9 + 16)/10 = 3 M1 A1
  3. E(X²) = (1 + 8 + 27 + 64)/10 = 10 ⇒ Var(X) = 10 − 9 = 1 M1 A1
Common mistakeVar(X) = Σx²p with the mean not subtracted, or squaring the probabilities instead of the x-values.

2The binomial distribution

X ~ B(n, p)

P(X = r) = ⁿCᵣ pʳ (1 − p)n − r ; E(X) = np ; Var(X) = np(1 − p) FORMULA LIST (MF19)
r = 0, 1, …, n
xP0123456789100.10.2X ~ B(10, 0.3), mean np = 3red: P(X ≤ 2) = 0.383tallest bar at x = 3
B(10, 0.3): the peak is near the mean np = 3.
  • Conditions in context: "each item is faulty independently with the same probability". Questions often ask you to state one.
Worked example

X ~ B(10, 0.3). Find (a) P(X = 3), (b) P(X ≤ 2), (c) P(X ≥ 2).

  1. (a) ¹⁰C₃ (0.3)³(0.7)⁷ = 0.267 M1 A1
  2. (b) P(0) + P(1) + P(2) = 0.0282 + 0.1211 + 0.2335 = 0.383 M1 A1
  3. (c) 1 − P(0) − P(1) = 0.851 M1 A1
Common mistake"At least 2" read as 1 − P(X ≤ 2). "At least 2" means X ≥ 2, so subtract P(X ≤ 1). Write the inequality before calculating.

3The geometric distribution

X ~ Geo(p)

P(X = r) = p(1 − p)r − 1 ; P(X > r) = (1 − p)ʳ ; E(X) = 1/p
r = 1, 2, 3, … (no upper limit)
  • Why P(X > r) = (1 − p)ʳ: the first r trials all fail.
Worked example

A fair die is thrown until a six appears. Find (a) P(X = 4), (b) P(X > 3), (c) E(X).

  1. (a) (5/6)³(1/6) = 0.0965 M1 A1
  2. (b) (5/6)³ = 0.579 B1
  3. (c) 6 B1
Worked example

Find the least number of throws needed for the probability of at least one six to exceed 0.9.

  1. 1 − (5/6)ⁿ > 0.9 ⇒ (5/6)ⁿ < 0.1 M1
  2. n > ln 0.1/ln(5/6) = 12.6 M1
  3. n = 13 A1
Common mistakeUsing binomial for "until the first success". The number of trials isn't fixed, so it's geometric.
Linking questionProbability rules are in Probability; ⁿCᵣ is in Permutations and combinations; large binomials are approximated in The normal distribution; and the Poisson distribution is in The Poisson distribution.

Test yourself — cover the answers first

Quick check

  1. X ~ B(20, 0.25). Find E(X).
  2. X ~ B(20, 0.25). Find Var(X).
  3. X ~ Geo(0.2). Find E(X).
  4. X ~ Geo(0.2). Find P(X = 2).
  5. State two conditions for a binomial model.

Exam-style

  1. X ~ B(8, 0.4). Find P(2 < X ≤ 5).
  2. The probability that a seed germinates is 0.85. 12 seeds are planted. Find the probability that at least 11 germinate.
  3. X ~ Geo(p) and P(X = 1) = 0.3. Find P(X ≥ 3).
Show answers
  1. 5
  2. 3.75
  3. 5
  4. 0.16
  5. fixed number of independent trials; constant probability of success
  6. P(3) + P(4) + P(5) = 0.2787 + 0.2322 + 0.1239 = 0.635
  7. 12(0.85)¹¹(0.15) + 0.85¹² = 0.3012 + 0.1422 = 0.443
  8. (0.7)² = 0.49
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.