Binomial and Poisson: 7 exam questions solved step by step

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What this lesson covers

  • Binomial counts successes in a fixed number of trials; Poisson counts events in an interval.
  • A Poisson rate scales with the interval, so a rate per hour becomes a rate per day by multiplying.
Key idea

7 questions, one method

  • 1. Name the distribution, and why.
  • 2. Turn the words into X ≤ or X ≥.
  • 3. Calculate, keeping 4 d.p.
  • 4. Round at the end, in context.
Your turn

Question 1 · Achieved

  • 85% of seeds in a packet germinate.
  • A gardener plants 12 of them.
  • Find P(exactly 10 germinate).
Worked example

Q1: X ~ Bin(12, 0.85)

  • Fixed n, two outcomes.
  • nCx × pˣ × (1 − p)ⁿ⁻ˣ
  1. X = seeds that germinate
  2. X ~ Bin(12, 0.85)fixed n = 12
  3. germinate or not, p fixedindependent seeds
  4. P(X = 10) = 12C10 × 0.85¹⁰ × 0.15²
  5. = 66 × 0.196874 × 0.0225
  6. = 0.292358
  7. = 0.29244 d.p.
  8. about a 29% chanceexactly 10 of 12
Your turn

Question 2 · Achieved

  • A fire station gets callouts at an
  • average rate of 2.5 a day.
  • Find P(exactly 4 callouts in a day).
Worked example

Q2: X ~ Po(2.5)

  • Events in an interval.
  • e^(−λ) × λˣ ÷ x!
  1. X = callouts in one day
  2. X ~ Po(2.5)mean rate 2.5 a day
  3. random, independentone at a time
  4. P(X = 4) = e^(−2.5) × 2.5⁴ ÷ 4!
  5. = 0.082085 × 39.0625 ÷ 24
  6. = 0.133602
  7. = 0.13364 d.p.
  8. about 1 day in 7 or 8exactly 4 callouts
Your turn

Question 3 · Merit

  • The same 12 seeds, 85% germinate.
  • Find P(at least 10 germinate).
  • Careful with the complement.
Worked example

Q3: at least 10

  • At least 10: 10, 11 or 12.
  • One minus "9 or fewer".
  1. X ~ Bin(12, 0.85)
  2. P(X ≥ 10) = 1 − P(X ≤ 9)complement
  3. 1 − P(X ≤ 10)loses X = 10
  4. P(X ≤ 9) = 0.2642cumulative
  5. P(X ≥ 10) = 1 − 0.2642
  6. = 0.7358
  7. 0.2924 + 0.3012 + 0.1422 = 0.7358
Your turn

Question 4 · Merit

  • The fire station: 2.5 callouts a day.
  • Find P(fewer than 15 callouts
  • in a 7-day week).
Worked example

Q4: scale the rate

  • Per day becomes per week.
  • Fewer than 15: up to 14.
  1. λ = 2.5 × 7 = 17.5a week
  2. X ~ Po(17.5)
  3. fewer than 15: X ≤ 14whole numbers
  4. P(X ≤ 14) = 0.242640
  5. = 0.24264 d.p.
  6. P(X ≤ 15) = 0.3275includes 15
  7. about a 24% chancea quiet week
Your turn

Question 5 · Merit, in context

  • 35% of a large Year 13 cohort have
  • a part-time job. In a random 20,
  • find P(5 to 9 of them have one),
  • 5 and 9 included.
Worked example

Q5: between 5 and 9

  • Up to 9, minus up to 4.
  • Keep 5 in.
  1. X ~ Bin(20, 0.35)large cohort: p fixed
  2. P(5 ≤ X ≤ 9)
  3. = P(X ≤ 9) − P(X ≤ 4)keeps X = 5
  4. P(X ≤ 9) = 0.8782
  5. P(X ≤ 4) = 0.1182
  6. 0.8782 − 0.1182 = 0.7600
  7. P(X ≤ 9) − P(X ≤ 5)loses X = 5
  8. about 76% of such groups
Your turn

Question 6 · Excellence

  • Knots occur at random in pine boards.
  • 45% of 1 m lengths have no knots.
  • Find the rate per metre, then
  • P(at least 2 knots in a 3 m board).
Worked example

Q6: the rate, backwards

  • No knots: P(X = 0).
  • Solve for λ with ln.
  1. X = knots in 1 m
  2. X ~ Po(λ)random along the wood
  3. P(X = 0) = 0.45
  4. e^(−λ) = 0.45P(0) = e^(−λ)
  5. −λ = ln 0.45
  6. λ = −ln 0.45 = 0.7985knots a metre
  7. e^(−0.7985) = 0.4500
Worked example

Q6: the 3 m board

  • Three times the length.
  • Three times the rate.
  1. λ = 0.7985 a metre
  2. λ = 3 × 0.7985 = 2.3955a 3 m board
  3. Y ~ Po(2.3955)
  4. P(Y ≥ 2) = 1 − P(Y ≤ 1)at least 2
  5. P(Y ≤ 1) = 0.3094
  6. 1 − 0.3094 = 0.6906
  7. about 69% of 3 m boards
Your turn

Question 7 · Excellence

  • A small plane has 68 seats. Each
  • ticket holder turns up with p = 0.92.
  • Keep P(overbooked) under 5%.
  • Can they sell 72? What is the most?
Worked example

Q7: how many tickets?

  • Overbooked: 69 or more.
  • Try n, step down.
  1. X ~ Bin(n, 0.92)X = who turns up
  2. overbooked: X ≥ 6968 seats
  3. P(X ≥ 69) = 1 − P(X ≤ 68)
  4. mean 72 × 0.92 = 66.24under 68 seats
  5. n = 72: 1 − 0.8375 = 0.1625over 0.05
  6. n = 71: 1 − 0.9303 = 0.0697over 0.05
  7. n = 70: 1 − 0.9793 = 0.0207under 0.05
  8. sell at most 70 tickets
  9. groups may not be independent
Summary

The method that solved all seven

  • Name the model, turn the words into inequalities, calculate, round at the end.
StepHow
Name itBin(n, p) or Po(λ), and why
At least1 − P(X ≤ one below)
Fewer thanX ≤ one below
Betweensubtract cumulatives
Ratescale λ to the interval
Backwardse^(−λ) = P(X = 0)

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