Hard Further Maths The Poisson distribution Questions

Challenging, exam-style Further Maths The Poisson distribution questions with worked solutions. Stretch yourself on the hardest poisson, cumulative-probability, inequalities, complement problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The random variable YY has the binomial distribution Y∼B(150,0.02)Y \sim \text{B}(150, 0.02). Which of the following is the exact value of P(Y≤5)P(Y \le 5), using the binomial distribution, correct to 44 decimal places?
Show worked solution

Worked solution

  1. Write down the exact distribution

    Y∼B(150,0.02)Y\sim\text{B}(150,0.02)

    The question asks for the exact binomial probability, not a Poisson approximation.

  2. Write down the binomial probability function

    P(Y=y)=(150y)(0.02)y(0.98)150−yP(Y=y)=\binom{150}{y}\left(0.02\right)^{y}\left(0.98\right)^{150-y}

    This is the exact model; the Poisson distribution would only approximate it.

  3. Evaluate P(Y=0)P(Y = 0)

    P(Y=0)=(1500)(0.02)0(0.98)150=0.048296P(Y=0)=\binom{150}{0}\left(0.02\right)^{0}\left(0.98\right)^{150}=0.048296

    Substitute y=0y=0 into the binomial probability function.

  4. Evaluate P(Y=1)P(Y = 1)

    P(Y=1)=(1501)(0.02)1(0.98)149=0.147845P(Y=1)=\binom{150}{1}\left(0.02\right)^{1}\left(0.98\right)^{149}=0.147845

    Substitute y=1y=1 into the binomial probability function.

  5. Evaluate P(Y=2)P(Y = 2)

    P(Y=2)=(1502)(0.02)2(0.98)148=0.224785P(Y=2)=\binom{150}{2}\left(0.02\right)^{2}\left(0.98\right)^{148}=0.224785

    Substitute y=2y=2 into the binomial probability function.

  6. Evaluate P(Y=3)P(Y = 3)

    P(Y=3)=(1503)(0.02)3(0.98)147=0.226314P(Y=3)=\binom{150}{3}\left(0.02\right)^{3}\left(0.98\right)^{147}=0.226314

    Substitute y=3y=3 into the binomial probability function.

  7. Evaluate P(Y=4)P(Y = 4)

    P(Y=4)=(1504)(0.02)4(0.98)146=0.169735P(Y=4)=\binom{150}{4}\left(0.02\right)^{4}\left(0.98\right)^{146}=0.169735

    Substitute y=4y=4 into the binomial probability function.

  8. Evaluate P(Y=5)P(Y = 5)

    P(Y=5)=(1505)(0.02)5(0.98)145=0.101148P(Y=5)=\binom{150}{5}\left(0.02\right)^{5}\left(0.98\right)^{145}=0.101148

    Substitute y=5y=5 into the binomial probability function.

  9. Add the terms

    P(Y≤5)=0.048296+0.147845+0.224785+0.226314+0.169735+0.101148=0.918123P(Y\le 5)=0.048296+0.147845+0.224785+0.226314+0.169735+0.101148=0.918123

    These are all the outcomes required.

  10. Round to the required accuracy

    P(Y≤5)=0.9181  (4 d.p.)P(Y \le 5)=0.9181\;(4\text{ d.p.})

    The unrounded exact value is 0.918123…0.918123\ldots.

  11. Compare with the Poisson approximation

    λ=np=3,P(Y≤5)≈0.9161  (Poisson)versus0.9181  (exact binomial)\lambda=np=3,\quad P(Y \le 5)\approx0.9161\;\text{(Poisson)}\quad\text{versus}\quad 0.9181\;\text{(exact binomial)}

    The Poisson approximation with λ=np=3\lambda=np=3 gives 0.91610.9161, compared with the exact binomial value 0.91810.9181, both to the stated accuracy. The Poisson value is only an estimate of the exact binomial probability.

  12. Recall the conditions for a Poisson model

    singly, independently, at a constant mean rate\text{singly, independently, at a constant mean rate}

    A Poisson model requires events to occur singly, independently and at a constant mean rate.

  13. Recall the mean and variance of a Poisson distribution

    E(X)=Var(X)=λ\text{E}(X)=\text{Var}(X)=\lambda

    For a Poisson distribution the mean and the variance are both equal to λ\lambda.

  14. Note that the probabilities of a Poisson distribution sum to 11

    ∑x=0∞e−λλxx!=1\sum_{x=0}^{\infty}\frac{e^{-\lambda}\lambda^{x}}{x!}=1

    The Poisson probabilities form a complete probability distribution.

  15. Select the matching option

    P(Y≤5)=0.9181P(Y \le 5)=0.9181

    This is the exact binomial probability.

Answer
P(Y≤5)=0.9181P(Y \le 5)=0.9181
Question 2
9 markschallenging
The random variables XX, YY and ZZ are independent, with X∼Po(2)X \sim \text{Po}(2), Y∼Po(1.5)Y \sim \text{Po}(1.5) and Z∼Po(1.5)Z \sim \text{Po}(1.5). The random variable TT is defined by T=X+Y+ZT = X + Y + Z. Which of the following is the value of P(T≥8)P(T \ge 8), correct to 44 decimal places?
Show worked solution

Worked solution

  1. Quote the additivity property of independent Poisson variables

    X+Y+Z∼Po(λ1+λ2+λ3)X+Y+Z\sim\text{Po}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)

    If the variables are independent, the sum of Poisson variables is Poisson with the sum of the means. Independence is stated in the question and is essential.

  2. Add the means

    λ=2+1.5+1.5=5\lambda=2+1.5+1.5=5

    The parameter of the combined distribution is the total of the separate means.

  3. Write down the distribution of TT

    T∼Po(5)T\sim\text{Po}(5)

    The total count is Poisson with mean λ=5\lambda=5.

  4. Write down the Poisson probability function

    P(T=x)=e−λλxx!,x=0,1,2,…P(T=x)=\frac{e^{-\lambda}\lambda^{x}}{x!},\quad x=0,1,2,\ldots

    This is the probability function of a Poisson distribution with mean λ\lambda.

  5. Convert the upper tail into a cumulative probability

    P(T≥8)=1−P(T≤7)P(T \ge 8)=1-P(T\le 7)

    A Poisson variable has no upper limit, so an upper tail is found as one minus a cumulative probability. Note that P(T≥8)P(T \ge 8) and P(T>8)P(T > 8) are different: here the terms x=0,…,7x=0,\ldots,7 are the ones subtracted.

  6. Evaluate P(T=0)P(T = 0)

    P(T=0)=e−5500!=0.006738P(T=0)=\frac{e^{-5}5^{0}}{0!}=0.006738

    Substitute x=0x=0 and λ=5\lambda=5 into the Poisson probability function.

  7. Evaluate P(T=1)P(T = 1)

    P(T=1)=e−5511!=0.033690P(T=1)=\frac{e^{-5}5^{1}}{1!}=0.033690

    Substitute x=1x=1 and λ=5\lambda=5 into the Poisson probability function.

  8. Evaluate P(T=2)P(T = 2)

    P(T=2)=e−5522!=0.084224P(T=2)=\frac{e^{-5}5^{2}}{2!}=0.084224

    Substitute x=2x=2 and λ=5\lambda=5 into the Poisson probability function.

  9. Evaluate P(T=3)P(T = 3)

    P(T=3)=e−5533!=0.140374P(T=3)=\frac{e^{-5}5^{3}}{3!}=0.140374

    Substitute x=3x=3 and λ=5\lambda=5 into the Poisson probability function.

  10. Evaluate P(T=4)P(T = 4)

    P(T=4)=e−5544!=0.175467P(T=4)=\frac{e^{-5}5^{4}}{4!}=0.175467

    Substitute x=4x=4 and λ=5\lambda=5 into the Poisson probability function.

  11. Evaluate P(T=5)P(T = 5)

    P(T=5)=e−5555!=0.175467P(T=5)=\frac{e^{-5}5^{5}}{5!}=0.175467

    Substitute x=5x=5 and λ=5\lambda=5 into the Poisson probability function.

  12. Evaluate P(T=6)P(T = 6)

    P(T=6)=e−5566!=0.146223P(T=6)=\frac{e^{-5}5^{6}}{6!}=0.146223

    Substitute x=6x=6 and λ=5\lambda=5 into the Poisson probability function.

  13. Evaluate P(T=7)P(T = 7)

    P(T=7)=e−5577!=0.104445P(T=7)=\frac{e^{-5}5^{7}}{7!}=0.104445

    Substitute x=7x=7 and λ=5\lambda=5 into the Poisson probability function.

  14. Add the probabilities

    P(T≤7)=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467+0.146223+0.104445=0.866628P(T\le 7)=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467+0.146223+0.104445=0.866628

    The individual terms are added. The inequality is not strict, so x=8x=8 is included in the tail.

  15. Subtract from 11

    P(T≥8)=1−0.866628=0.133372P(T \ge 8)=1-0.866628=0.133372

    The complement of the cumulative probability gives the upper tail.

  16. Round to the required accuracy

    P(T≥8)=0.1334  (4 d.p.)P(T \ge 8)=0.1334\;(4\text{ d.p.})

    The unrounded value is 0.133372…0.133372\ldots, which rounds to 0.13340.1334.

  17. Select the matching option

    P(T≥8)=0.1334P(T \ge 8)=0.1334

    This is the value of the required probability.

Answer
P(T≥8)=0.1334P(T \ge 8)=0.1334
Question 3
9 markschallenging
Complaints reach a call centre independently, singly and at a constant mean rate of 1.751.75 per day. The random variable XX denotes the number of complaints in a period of 44 days. Which of the following is the value of P(X≤6)P(X \le 6), correct to 44 decimal places?
Show worked solution

Worked solution

  1. Check the conditions for a Poisson model

    singly,  independently,  constant mean rate\text{singly},\;\text{independently},\;\text{constant mean rate}

    The three modelling conditions are stated in the question, so a Poisson model is appropriate.

  2. Scale the mean to the length of the interval

    λ=1.75×4=7\lambda=1.75\times4=7

    The mean is 1.751.75 per day, so over 44 days the mean is 77.

  3. Write down the distribution of XX

    X∼Po(7)X\sim\text{Po}(7)

    The count over the stated interval is Poisson with mean λ=7\lambda=7.

  4. Write down the Poisson probability function

    P(X=x)=e−λλxx!,x=0,1,2,…P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!},\quad x=0,1,2,\ldots

    This is the probability function of a Poisson distribution with mean λ\lambda.

  5. Evaluate P(X=0)P(X = 0)

    P(X=0)=e−7700!=0.000912P(X=0)=\frac{e^{-7}7^{0}}{0!}=0.000912

    Substitute x=0x=0 and λ=7\lambda=7 into the Poisson probability function.

  6. Evaluate P(X=1)P(X = 1)

    P(X=1)=e−7711!=0.006383P(X=1)=\frac{e^{-7}7^{1}}{1!}=0.006383

    Substitute x=1x=1 and λ=7\lambda=7 into the Poisson probability function.

  7. Evaluate P(X=2)P(X = 2)

    P(X=2)=e−7722!=0.022341P(X=2)=\frac{e^{-7}7^{2}}{2!}=0.022341

    Substitute x=2x=2 and λ=7\lambda=7 into the Poisson probability function.

  8. Evaluate P(X=3)P(X = 3)

    P(X=3)=e−7733!=0.052129P(X=3)=\frac{e^{-7}7^{3}}{3!}=0.052129

    Substitute x=3x=3 and λ=7\lambda=7 into the Poisson probability function.

  9. Evaluate P(X=4)P(X = 4)

    P(X=4)=e−7744!=0.091226P(X=4)=\frac{e^{-7}7^{4}}{4!}=0.091226

    Substitute x=4x=4 and λ=7\lambda=7 into the Poisson probability function.

  10. Evaluate P(X=5)P(X = 5)

    P(X=5)=e−7755!=0.127717P(X=5)=\frac{e^{-7}7^{5}}{5!}=0.127717

    Substitute x=5x=5 and λ=7\lambda=7 into the Poisson probability function.

  11. Evaluate P(X=6)P(X = 6)

    P(X=6)=e−7766!=0.149003P(X=6)=\frac{e^{-7}7^{6}}{6!}=0.149003

    Substitute x=6x=6 and λ=7\lambda=7 into the Poisson probability function.

  12. Add the probabilities

    P(X≤6)=0.000912+0.006383+0.022341+0.052129+0.091226+0.127717+0.149003=0.449711P(X\le 6)=0.000912+0.006383+0.022341+0.052129+0.091226+0.127717+0.149003=0.449711

    The individual terms are added. Here P(X≤6)P(X \le 6) includes x=6x=6, whereas P(X<6)P(X < 6) would not.

  13. Round to the required accuracy

    P(X≤6)=0.4497  (4 d.p.)P(X \le 6)=0.4497\;(4\text{ d.p.})

    The unrounded value is 0.449711…0.449711\ldots, which rounds to 0.44970.4497.

  14. Recall the conditions for a Poisson model

    singly, independently, at a constant mean rate\text{singly, independently, at a constant mean rate}

    A Poisson model requires events to occur singly, independently and at a constant mean rate.

  15. Select the matching option

    P(X≤6)=0.4497P(X \le 6)=0.4497

    This is the value of the required probability.

Answer
P(X≤6)=0.4497P(X \le 6)=0.4497
Question 4
9 markschallenging
The random variable XX has a Poisson distribution with mean λ\lambda, where λ>0\lambda > 0. It is given that P(X=2)=2P(X=1)P(X = 2) = 2P(X = 1). Find the value of λ\lambda and hence find P(2≤X≤6)P(2 \le X \le 6), giving your answer to 44 decimal places.
Show worked solution

Worked solution

  1. Write both probabilities using the Poisson probability function

    e−λλ22!=2⋅e−λλ11!\frac{e^{-\lambda}\lambda^{2}}{2!}=2\cdot\frac{e^{-\lambda}\lambda^{1}}{1!}

    Both sides use P(X=x)=e−λλx/x!P(X=x)=e^{-\lambda}\lambda^{x}/x!.

  2. Cancel the common factor e−λe^{-\lambda} and the lower power of λ\lambda

    λ12=2⋅11\frac{\lambda^{1}}{2}=2\cdot\frac{1}{1}

    Since λ>0\lambda>0 the factors e−λe^{-\lambda} and λ1\lambda^{1} may be divided out.

  3. Rearrange to find λ1\lambda^{1}

    λ1=2×21=4\lambda^{1}=2\times\frac{2}{1}=4

    Multiplying up leaves a single power of λ\lambda.

  4. Take the positive root

    λ=4\lambda=4

    The mean of a Poisson distribution must be positive.

  5. Write down the distribution

    X∼Po(4)X\sim\text{Po}(4)

    The value of λ\lambda found above now fixes the distribution.

  6. Write down the Poisson probability function

    P(X=x)=e−λλxx!,x=0,1,2,…P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!},\quad x=0,1,2,\ldots

    This is the probability function of a Poisson distribution with mean λ\lambda.

  7. Identify exactly which values of xx are included

    P(2≤X≤6)=∑x=26e−λλxx!P(2 \le X \le 6)=\sum_{x=2}^{6}\frac{e^{-\lambda}\lambda^{x}}{x!}

    Read the inequality signs carefully: the values x=2,…,6x=2,\ldots,6 are included, and no others.

  8. Evaluate P(X=2)P(X = 2)

    P(X=2)=e−4422!=0.146525P(X=2)=\frac{e^{-4}4^{2}}{2!}=0.146525

    Substitute x=2x=2 and λ=4\lambda=4 into the Poisson probability function.

  9. Evaluate P(X=3)P(X = 3)

    P(X=3)=e−4433!=0.195367P(X=3)=\frac{e^{-4}4^{3}}{3!}=0.195367

    Substitute x=3x=3 and λ=4\lambda=4 into the Poisson probability function.

  10. Evaluate P(X=4)P(X = 4)

    P(X=4)=e−4444!=0.195367P(X=4)=\frac{e^{-4}4^{4}}{4!}=0.195367

    Substitute x=4x=4 and λ=4\lambda=4 into the Poisson probability function.

  11. Evaluate P(X=5)P(X = 5)

    P(X=5)=e−4455!=0.156293P(X=5)=\frac{e^{-4}4^{5}}{5!}=0.156293

    Substitute x=5x=5 and λ=4\lambda=4 into the Poisson probability function.

  12. Evaluate P(X=6)P(X = 6)

    P(X=6)=e−4466!=0.104196P(X=6)=\frac{e^{-4}4^{6}}{6!}=0.104196

    Substitute x=6x=6 and λ=4\lambda=4 into the Poisson probability function.

  13. Add the probabilities

    P(2≤X≤6)=0.146525+0.195367+0.195367+0.156293+0.104196=0.797748P(2\le X\le 6)=0.146525+0.195367+0.195367+0.156293+0.104196=0.797748

    The individual terms are added.

  14. Round to the required accuracy

    P(2≤X≤6)=0.7977  (4 d.p.)P(2 \le X \le 6)=0.7977\;(4\text{ d.p.})

    The unrounded value is 0.797748…0.797748\ldots.

  15. State the final answer

    P(2≤X≤6)=0.7977P(2 \le X \le 6)=0.7977

    This is the required probability.

Answer
P(2≤X≤6)=0.7977P(2 \le X \le 6)=0.7977
Question 5
9 markschallenging
Lorries arrive at a distribution depot independently, singly and at a constant mean rate of 44 per hour. The random variable XX denotes the number of lorries in a period of 33 hours. Find P(8≤X≤12)P(8 \le X \le 12), giving your answer to 44 decimal places.
Show worked solution

Worked solution

  1. Check the conditions for a Poisson model

    singly,  independently,  constant mean rate\text{singly},\;\text{independently},\;\text{constant mean rate}

    The three modelling conditions are stated in the question, so a Poisson model is appropriate.

  2. Scale the mean to the length of the interval

    λ=4×3=12\lambda=4\times3=12

    The mean is 44 per hour, so over 33 hours the mean is 1212.

  3. Write down the distribution of XX

    X∼Po(12)X\sim\text{Po}(12)

    The count over the stated interval is Poisson with mean λ=12\lambda=12.

  4. Write down the Poisson probability function

    P(X=x)=e−λλxx!,x=0,1,2,…P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!},\quad x=0,1,2,\ldots

    This is the probability function of a Poisson distribution with mean λ\lambda.

  5. Identify exactly which values of xx are included

    P(8≤X≤12)=∑x=812e−λλxx!P(8 \le X \le 12)=\sum_{x=8}^{12}\frac{e^{-\lambda}\lambda^{x}}{x!}

    Read the inequality signs carefully: the values x=8,…,12x=8,\ldots,12 are included, and no others.

  6. Evaluate P(X=8)P(X = 8)

    P(X=8)=e−121288!=0.065523P(X=8)=\frac{e^{-12}12^{8}}{8!}=0.065523

    Substitute x=8x=8 and λ=12\lambda=12 into the Poisson probability function.

  7. Evaluate P(X=9)P(X = 9)

    P(X=9)=e−121299!=0.087364P(X=9)=\frac{e^{-12}12^{9}}{9!}=0.087364

    Substitute x=9x=9 and λ=12\lambda=12 into the Poisson probability function.

  8. Evaluate P(X=10)P(X = 10)

    P(X=10)=e−12121010!=0.104837P(X=10)=\frac{e^{-12}12^{10}}{10!}=0.104837

    Substitute x=10x=10 and λ=12\lambda=12 into the Poisson probability function.

  9. Evaluate P(X=11)P(X = 11)

    P(X=11)=e−12121111!=0.114368P(X=11)=\frac{e^{-12}12^{11}}{11!}=0.114368

    Substitute x=11x=11 and λ=12\lambda=12 into the Poisson probability function.

  10. Evaluate P(X=12)P(X = 12)

    P(X=12)=e−12121212!=0.114368P(X=12)=\frac{e^{-12}12^{12}}{12!}=0.114368

    Substitute x=12x=12 and λ=12\lambda=12 into the Poisson probability function.

  11. Add the probabilities

    P(8≤X≤12)=0.065523+0.087364+0.104837+0.114368+0.114368=0.486461P(8\le X\le 12)=0.065523+0.087364+0.104837+0.114368+0.114368=0.486461

    The individual terms are added.

  12. Round to the required accuracy

    P(8≤X≤12)=0.4865  (4 d.p.)P(8 \le X \le 12)=0.4865\;(4\text{ d.p.})

    The unrounded value is 0.486461…0.486461\ldots, which rounds to 0.48650.4865.

  13. Recall the conditions for a Poisson model

    singly, independently, at a constant mean rate\text{singly, independently, at a constant mean rate}

    A Poisson model requires events to occur singly, independently and at a constant mean rate.

  14. Recall the mean and variance of a Poisson distribution

    E(X)=Var(X)=λ\text{E}(X)=\text{Var}(X)=\lambda

    For a Poisson distribution the mean and the variance are both equal to λ\lambda.

  15. State the final answer

    P(8≤X≤12)=0.4865P(8 \le X \le 12)=0.4865

    This is the required probability, correct to the stated accuracy.

Answer
P(8≤X≤12)=0.4865P(8 \le X \le 12)=0.4865

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