FS1-2 Poisson and binomial distributions — revision question pack

3 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FS1-2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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FS1-2.1 · The Poisson distribution. The additive property of Poisson distributions.

Explanation

A Poisson distribution models a count of events occurring independently at a constant mean rate: $P(X=x)=e^{-\lambda}\dfrac{\lambda^x}{x!}$ for $x=0,1,2,\ldots$ . Scale $\lambda$ in proportion to the length or size of the observation interval.
If $X\sim\operatorname{Po}(\lambda)$ and $Y\sim\operatorname{Po}(\mu)$ are independent, then $X+Y\sim\operatorname{Po}(\lambda+\mu)$ .
This additive property also combines disjoint intervals of a Poisson process.
A real-world answer may require critical comment on constant rate and independence.
Examiners expect the parameter to match the requested interval and cumulative probabilities to use the correct complement and inclusive boundary.

A typical discrete, right-skewed Poisson probability mass function.

Worked example

Calls arrive at mean rate $1.4$ per $10$ minutes. Find the probability of at least $5$ calls in $30$ minutes.

1.For $30$ minutes, $\lambda=3(1.4)=4.2$ .
2.Let $X\sim\operatorname{Po}(4.2)$ .
3. $P(X\geq5)=1-P(X\leq4)=0.4102$ to $4$ significant figures.

Answer: The probability is $0.4102$ .

Common mistakes

Don't use the rate for $10$ minutes as the parameter for a $30$ -minute count.
Don't add parameters for counts that are not independent.
Don't calculate $1-P(X\leq5)$ for the event $X\geq5$ .

Exam tip

Write the scaled distribution before using a calculator; this makes the interval conversion explicit.

Tier 1 · Easy

1.
A sensor records a Poisson-distributed number of amber flashes with mean $2.7$ per minute. Find the probability of no amber flashes in one minute.
(2)
(Total for Question 1 is 2 marks)
2.
Events in the first $15$ minutes and in the final $21$ minutes of one hour are recorded separately. They are modelled by a Poisson process with mean rate $5$ per hour. State the distribution of each count. A fault makes events occur in pairs one minute apart. Identify the Poisson-process assumption that this feature breaks.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Independent counts $A$ and $B$ have distributions $\operatorname{Po}(3.2)$ and $\operatorname{Po}(1.7)$ . Find $P(A+B\geq7)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Requests reach services A and B at mean rates $r$ and $1.6$ per hour. The combined count over $2.5$ hours has distribution $\operatorname{Po}(13.5)$ . Find $r$ , then find the probability of exactly $4$ requests reaching service A in half an hour. State the assumptions needed to combine the two streams.
(5)
(Total for Question 2 is 5 marks)
3.
Messages follow a Poisson process at a mean rate of $4.5$ per hour. Counts $A$ , $B$ and $C$ are recorded in three non-overlapping windows lasting $8$ , $20$ and $32$ minutes respectively. Find the probability that between three and five messages inclusive are recorded across the three windows in total.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Inspection pings are modelled by a Poisson process at a mean rate of $0.8$ per $10$ minutes. Find the probability of exactly $2$ pings in the first $15$ minutes and at most $2$ pings in the next $30$ minutes. State two assumptions needed for the model and one feature of the real process that would make it unsuitable.
(8)
(Total for Question 1 is 8 marks)
2.
Events follow a Poisson process at rate $6$ per hour. Let $X$ count events from minute $0$ to minute $30$ , and let $Y$ count events from minute $20$ to minute $50$ . Show that the number $Z$ of events in the union from minute $0$ to minute $50$ has distribution $\operatorname{Po}(5)$ , find $P(Z=2)$ , and explain why the additive property for independent Poisson variables cannot be used to find $\operatorname{Var}(X+Y)$ .
(6)
(Total for Question 2 is 6 marks)
3.
Defects along a cable are modelled by a Poisson process with mean rate $1.4$ per kilometre. Find the length $L$ of cable for which $P(\text{no defects in }L)=0.2$ . Then find $P(\text{at least 2 defects in }1.5\text{ km})$ . State two assumptions needed for the Poisson-process model.
(8)
(Total for Question 3 is 8 marks)
4.
Two independent Poisson processes have mean rates $2.4$ and $1.6$ events per hour respectively. Let $A$ and $B$ be their respective event counts during a $1.5$ -hour period. Given that $A+B=5$ , find the conditional probability that $A=3$ , and the conditional probability that $A>B$ .
(7)
(Total for Question 4 is 7 marks)
5.
Arrivals follow a Poisson process with mean rate $r$ per hour. The probability of at most one arrival in a $30$ -minute period is $0.7$ . Find $r$ to $5$ significant figures. Over a later $2$ -hour period, an independent background source contributes a Poisson number of arrivals with mean $1.2$ . Find the probability that the combined number of arrivals is at least $8$ , giving your answer to $4$ decimal places.
(8)
(Total for Question 5 is 8 marks)

FS1-2.2 · The mean and variance of the binomial distribution and the Poisson distribution.

Explanation

For $X\sim\operatorname{Bin}(n,p)$ , $E(X)=np$ and $\operatorname{Var}(X)=np(1-p)$ . For $Y\sim\operatorname{Po}(\lambda)$ , both mean and variance equal $\lambda$ .
Under a linear transformation, $E(aX+b)=aE(X)+b$ but $\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$ because adding a constant changes location, not spread.
These relationships can recover unknown distribution parameters from given means or variances and help compare models with observed data.
The binomial parameter $n$ must be a positive integer and $0\leq p\leq1$ .
Examiners expect simultaneous parameter equations to be solved exactly where possible and a standard deviation to be the positive square root of variance.

Worked example

A binomial random variable has mean $24$ and variance $18$ . Find $n$ and $p$ .

1. $np=24$ and $np(1-p)=18$ .
2.Divide the variance equation by the mean equation: $1-p=\dfrac{18}{24}=\dfrac34$ .
3.Thus $p=\dfrac14$ and $n=24/(1/4)=96$ .

Answer: $n=96$ and $p=\dfrac14$ .

Common mistakes

Don't use $np$ as both the mean and variance of a binomial distribution.
Don't add $b$ to the variance when transforming $aX+b$ .
Don't multiply variance by $a$ rather than $a^2$ .

Exam tip

When both binomial mean and variance are given, divide the equations first to isolate $1-p$ .

Tier 1 · Easy

1.
Given $X\sim\operatorname{Bin}(80,0.15)$ , find $E(X)$ and $\operatorname{Var}(X)$ .
(2)
(Total for Question 1 is 2 marks)
2.
A count variable has mean $6.25$ and variance $6.25$ . It is to be modelled as either Poisson or binomial. State, giving a reason, which of these models is consistent with these summaries. Find its standard deviation.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The random variable $Y$ is Poisson and $\operatorname{Var}(2Y+3)=28$ . Find the parameter of $Y$ and $E(2Y+3)$ .
(4)
(Total for Question 1 is 4 marks)
2.
Let $X\sim\operatorname{Bin}(n,p)$ and define $Y=5-2X$ . Given that $E(Y)=-11$ and $\operatorname{Var}(Y)=19.2$ , find $n$ and $p$ . Verify that these values of $n$ and $p$ are possible for a binomial distribution.
(6)
(Total for Question 2 is 6 marks)
3.
Let $X\sim\operatorname{Bin}(n,p)$ . Given that $E(X)=9$ and $E(X^2)=88.2$ , determine $n$ and $p$ , then find $P(X=0)$ to $4$ significant figures.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
A binomial random variable $X$ has mean $18$ and variance $13.5$ . Determine $n$ and $p$ . Hence find $P(X=18)$ , giving your answer to $4$ significant figures.
(6)
(Total for Question 1 is 6 marks)
2.
Given $E(3X-2)=34$ and $\operatorname{Var}(3X-2)=27$ , find $n$ and $p$ for $X\sim\operatorname{Bin}(n,p)$ , and hence find $P(X\geq15)$ .
(7)
(Total for Question 2 is 7 marks)
3.
Let $X\sim\operatorname{Bin}(50,p)$ . Given that $\operatorname{Var}(X)=8$ , find the two possible values of $p$ . It is also known that $E(X)>25$ . Hence determine $p$ , $E(X)$ and the standard deviation of $X$ , and find $P(38\leq X\leq40)$ to $4$ significant figures.
(7)
(Total for Question 3 is 7 marks)
4.
Let $X\sim\operatorname{Bin}(n,0.35)$ . Given that $\operatorname{Var}(X)=9.1$ , determine $n$ . Find $E(X)$ , then find the mean and variance of $W=4X-7$ . Hence find $P(41\leq W\leq57)$ , giving your answer to $4$ decimal places.
(8)
(Total for Question 4 is 8 marks)
5.
Let $X\sim\operatorname{Bin}(n,p)$ , where $E(X)=9$ and $P(X=n)=0.1P(X=n-1)$ . Determine $n$ and $p$ , then find $\operatorname{Var}(X)$ and $P(X\geq n-1)$ , giving the probability to $6$ decimal places.
(8)
(Total for Question 5 is 8 marks)

FS1-2.3 · The use of the Poisson distribution as an approximation to the binomial distribution.

Explanation

When $n$ is large and $p$ is small, $X\sim\operatorname{Bin}(n,p)$ may be approximated by $Y\sim\operatorname{Po}(np)$ . State why the conditions are reasonable and set $\lambda=np$ .
The event boundary remains discrete and unchanged: no continuity correction is used for a binomial-to-Poisson approximation.
Calculator cumulative functions or complements can then evaluate the probability.
The approximation becomes less reliable when $p$ is not small or when the binomial trials are not independent with constant success probability.
Examiners expect approximation notation or clear wording, the value of $\lambda$ , and a probability answer labelled approximate rather than exact.

Worked example

Let $X\sim\operatorname{Bin}(800,0.006)$ . Use a Poisson approximation to estimate $P(X\geq7)$ .

1. $n$ is large, $p$ is small and $np=4.8$ , so use $Y\sim\operatorname{Po}(4.8)$ .
2. $P(X\geq7)\approx P(Y\geq7)$ .
3. $P(Y\geq7)=1-P(Y\leq6)=0.2092$ to $4$ significant figures.

Answer: $P(X\geq7)\approx0.2092$ .

Common mistakes

Don't use $\lambda=p$ instead of $\lambda=np$ .
Don't apply a continuity correction even though both distributions are discrete.
Don't quote the approximation without checking that $n$ is large and $p$ is small.

Exam tip

A full approximation line should state the conditions, $Y\sim\operatorname{Po}(np)$ and the unchanged event.

Tier 1 · Easy

1.
Let $X\sim\operatorname{Bin}(900,0.004)$ . State a suitable Poisson approximation and use it to estimate $P(X=0)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Let $X\sim\operatorname{Bin}(1200,0.0025)$ . Using a suitable Poisson approximation, find the smallest integer $k$ for which $P(X\leq k)\geq0.9$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The number of flawed seals in a batch has distribution $X\sim\operatorname{Bin}(400,0.01)$ . Use a Poisson approximation to estimate $P(X\geq6)$ .
(4)
(Total for Question 1 is 4 marks)
2.
Let $X\sim\operatorname{Bin}(2000,0.002)$ . Use a Poisson approximation to find $P(3\leq X\leq6)$ . Explain why the approximation is valid here but would not be valid if the success probability were $0.2$ .
(5)
(Total for Question 2 is 5 marks)
3.
Components are flawed independently with probability $0.003$ . An inspector chooses a sample size $n$ so that the expected number $X$ of flawed components is $5.4$ . Find $n$ , then use a suitable Poisson approximation to estimate $P(4\leq X\leq7)$ , giving your answer to $4$ decimal places.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
For $X\sim\operatorname{Bin}(200,0.015)$ , calculate $P(X\leq2)$ exactly and by a Poisson approximation. Find the percentage error of the approximation relative to the exact value.
(6)
(Total for Question 1 is 6 marks)
2.
Independent items fail with probability $0.004$ . Use a Poisson approximation to find the smallest sample size $n$ for which the probability of at most one failure is no greater than $0.1$ .
(6)
(Total for Question 2 is 6 marks)
3.
A batch contains $500$ independently produced components, each having the same small probability $p$ of a flaw. For a Poisson approximation $Y$ , it is known that $P(Y=3)/P(Y=2)=1.4$ . Estimate $p$ , then estimate the probability that at least six components are flawed, giving your answer to $4$ decimal places. Explain why a Poisson approximation is reasonable here.
(6)
(Total for Question 3 is 6 marks)
4.
Two independent batches contain $800$ and $500$ components. Components in the first batch are flawed independently with probability $0.003$ , and components in the second are flawed independently with probability $0.007$ . Let $X$ and $Y$ be the respective numbers of flawed components. Use Poisson approximations to estimate $P(X\leq1\text{ and }X+Y\geq5)$ , giving your answer to $4$ decimal places.
(7)
(Total for Question 4 is 7 marks)
5.
Three independent batches each contain $n$ components, and every component is flawed independently with probability $0.002$ . Use Poisson approximations to find the largest integer $n$ for which the probability that at least one batch contains no flawed components is at least $0.5$ .
(8)
(Total for Question 5 is 8 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FS1-2.1 · The Poisson distribution. The additive property of Poisson distributions.

Tier 1 · Easy

Mark scheme for FS1-2.1 Tier 1 · Easy
Question	Scheme	Marks
1	$P(X=0)=0.0672$ to $4$ significant figures	2
(2 marks)		2
Notes
With $X\sim\operatorname{Po}(2.7)$ , $P(X=0)=e^{-2.7}\dfrac{2.7^0}{0!}=e^{-2.7}=0.0672055\ldots$ .
2	First count $\operatorname{Po}(1.25)$ Second count $\operatorname{Po}(1.75)$ Events do not occur independently, because each event is accompanied by a second event one minute later	3
(3 marks)		3
Notes
Scale the hourly mean to each interval: $5(15/60)=1.25$ and $5(21/60)=1.75$ . A Poisson process assumes that events occur independently. Here each event is accompanied by a second event one minute later, so that assumption fails.

Tier 2 · Standard

Mark scheme for FS1-2.1 Tier 2 · Standard
Question	Scheme	Marks
1	$P(A+B\geq7)=0.2233$ to $4$ significant figures	3
(3 marks)		3
Notes
Independence gives $A+B\sim\operatorname{Po}(4.9)$ . Hence $P(A+B\geq7)=1-P(A+B\leq6)=0.223345\ldots$ .
2	$2.5(r+1.6)=13.5$ $r=3.8$ per hour The half-hour parameter for A is $1.9$ $P(N=4)=e^{-1.9}\dfrac{1.9^4}{4!}=0.0812$ to $3$ significant figures The calculation relies on constant rates and independence of the two services	5
(5 marks)		5
Notes
The total parameter is exposure multiplied by the sum of the independent rates: $2.5(r+1.6)=13.5$ , giving $r=3.8$ per hour. For half an hour the A-count $N\sim\operatorname{Po}(0.5\times3.8)=\operatorname{Po}(1.9)$ , so $P(N=4)=e^{-1.9}1.9^4/4!$ . The decomposition would fail if either rate changed during observation or the two streams were dependent.
3	$A\sim\operatorname{Po}(0.6)$ $B\sim\operatorname{Po}(1.5)$ $C\sim\operatorname{Po}(2.4)$ $A+B+C\sim\operatorname{Po}(4.5)$ $P(3\leq A+B+C\leq5)=\displaystyle\sum_{x=3}^{5}e^{-4.5}\dfrac{4.5^x}{x!}=0.5294$ to $4$ significant figures	5
(5 marks)		5
Notes
Scaling the hourly rate gives parameters $4.5(8/60)=0.6$ , $4.5(20/60)=1.5$ and $4.5(32/60)=2.4$ . Counts on the non-overlapping windows are independent, so their sum is $\operatorname{Po}(0.6+1.5+2.4)=\operatorname{Po}(4.5)$ . Therefore the required probability is $\sum_{x=3}^{5}e^{-4.5}4.5^x/x!=0.5293523639\ldots=0.5294$ to $4$ significant figures.

Tier 3 · Hard

Mark scheme for FS1-2.1 Tier 3 · Hard
Question	Scheme	Marks
1	$0.1235$ to $4$ significant figures Pings occur independently and at a constant mean rate For example, fault-triggered clustering or a changing rate during maintenance would make the model unsuitable	8
(8 marks)		8
Notes
The disjoint intervals have independent counts $X\sim\operatorname{Po}(1.2)$ and $Y\sim\operatorname{Po}(2.4)$ . Therefore the probability is $P(X=2)P(Y\leq2)=\left(e^{-1.2}\dfrac{1.2^2}{2!}\right)\left(e^{-2.4}\sum_{r=0}^{2}\dfrac{2.4^r}{r!}\right)=0.1235469\ldots$ . A Poisson process also requires independent events and a constant mean rate. Clustering after a fault would violate independence, while maintenance cycles could change the rate.
2	Use disjoint $20$ -, $10$ - and $20$ -minute counts $A,B,C$ $A,C\sim\operatorname{Po}(2)$ and $B\sim\operatorname{Po}(1)$ independently $Z=A+B+C\sim\operatorname{Po}(5)$ $P(Z=2)=e^{-5}\dfrac{5^2}{2!}=0.0842$ to $3$ significant figures $X$ and $Y$ share the overlap count $B$ and are not independent Therefore the independent-Poisson additive property does not apply to $X+Y$	6
(6 marks)		6
Notes
Partition the union into counts $A,B,C$ on minutes $0$ -- $20$ , $20$ -- $30$ and $30$ -- $50$ . They are independent with parameters $2,1,2$ , so their sum $Z\sim\operatorname{Po}(5)$ and $P(Z=2)=e^{-5}5^2/2!=0.0842243\ldots$ . However, $X=A+B$ and $Y=B+C$ share the random count $B$ , so $X$ and $Y$ are dependent and the additive property for independent Poisson variables cannot determine $\operatorname{Var}(X+Y)$ .
3	The number of defects in length $L$ has distribution $\operatorname{Po}(1.4L)$ $e^{-1.4L}=0.2$ $L=\dfrac{\ln5}{1.4}=1.15$ km to $3$ significant figures The number of defects in $1.5$ km has distribution $\operatorname{Po}(2.1)$ $P(N\geq2)=1-P(N=0)-P(N=1)=1-e^{-2.1}(1+2.1)$ $P(N\geq2)=0.6204$ to $4$ significant figures For example, the mean defect rate is constant per unit length Counts on disjoint lengths are independent	8
(8 marks)		8
Notes
For a length $L$ km the parameter is $1.4L$ , so $P(N=0)=e^{-1.4L}=0.2$ . Taking logarithms gives $-1.4L=\ln(0.2)=-\ln5$ , hence $L=\ln5/1.4=1.1495985089\ldots$ km. For $1.5$ km the parameter is $1.4(1.5)=2.1$ , so $P(N\geq2)=1-e^{-2.1}-2.1e^{-2.1}=1-e^{-2.1}(1+2.1)=0.6203850724\ldots$ . The model assumes that the mean defect rate remains constant per unit length and that counts on disjoint lengths are independent.
4	$A\sim\operatorname{Po}(3.6)$ and $B\sim\operatorname{Po}(2.4)$ $T=A+B\sim\operatorname{Po}(6)$ $P(A=a,B=5-a)=e^{-6}\dfrac{3.6^a2.4^{5-a}}{a!(5-a)!}$ $P(A=a\mid T=5)=\binom5a(0.6)^a(0.4)^{5-a}$ $P(A=3\mid T=5)=0.3456$ $A>B$ is equivalent to $A\geq3$ when $T=5$ $P(A>B\mid T=5)=0.68256$	7
(7 marks)		7
Notes
Over $1.5$ hours the independent counts have parameters $3.6$ and $2.4$ , so their total $T\sim\operatorname{Po}(6)$ . Independence gives $P(A=a,B=5-a)=e^{-6}3.6^a2.4^{5-a}/[a!(5-a)!]$ . Dividing by $P(T=5)=e^{-6}6^5/5!$ gives $P(A=a\mid T=5)=\binom5a(0.6)^a(0.4)^{5-a}$ . Hence $P(A=3\mid T=5)=0.3456$ . Since $B=5-A$ , the event $A>B$ is $A\geq3$ , whose conditional probability is $\sum_{a=3}^{5}\binom5a(0.6)^a(0.4)^{5-a}=0.68256$ .
5	For $30$ minutes, $N\sim\operatorname{Po}(r/2)$ $e^{-r/2}(1+r/2)=0.7$ $r/2=1.097349\ldots$ $r=2.1947$ per hour to $5$ significant figures The main-source count over $2$ hours has parameter $2r$ The combined count has distribution $\operatorname{Po}(2r+1.2)$ $P(T\geq8)=1-\displaystyle\sum_{t=0}^{7}e^{-(2r+1.2)}\dfrac{(2r+1.2)^t}{t!}$ $P(T\geq8)=0.2017$ to $4$ decimal places	8
(8 marks)		8
Notes
In $30$ minutes the Poisson parameter is $r/2$ , so the given probability yields $e^{-r/2}(1+r/2)=0.7$ . Numerical solution gives $r/2=1.0973492107\ldots$ and hence $r=2.1946984214\ldots$ per hour. Over $2$ hours the main source has parameter $2r$ , and independence allows the background parameter $1.2$ to be added. Therefore $T\sim\operatorname{Po}(2r+1.2)=\operatorname{Po}(5.5893968428\ldots)$ and $P(T\geq8)=1-P(T\leq7)=0.2016828938\ldots=0.2017$ .

FS1-2.2 · The mean and variance of the binomial distribution and the Poisson distribution.

Tier 1 · Easy

Mark scheme for FS1-2.2 Tier 1 · Easy
Question	Scheme	Marks
1	$E(X)=12$ $\operatorname{Var}(X)=10.2$	2
(2 marks)		2
Notes
$E(X)=np=80(0.15)=12$ . Also $\operatorname{Var}(X)=np(1-p)=80(0.15)(0.85)=10.2$ .
2	A Poisson model is consistent because its mean and variance are equal The Poisson parameter is $6.25$ The standard deviation is the positive root $\sqrt{6.25}=2.5$	3
(3 marks)		3
Notes
A Poisson variable has equal mean and variance, so $X\sim\operatorname{Po}(6.25)$ is consistent. A non-degenerate binomial variable has variance $np(1-p)$ strictly below its mean $np$ , so it is inconsistent. Standard deviation is non-negative, hence $\sqrt{6.25}=2.5$ .

Tier 2 · Standard

Mark scheme for FS1-2.2 Tier 2 · Standard
Question	Scheme	Marks
1	$Y\sim\operatorname{Po}(7)$ $E(2Y+3)=17$	4
(4 marks)		4
Notes
If $Y\sim\operatorname{Po}(\lambda)$ , then $\operatorname{Var}(2Y+3)=4\lambda$ . Hence $4\lambda=28$ and $\lambda=7$ . Therefore $E(2Y+3)=2E(Y)+3=2(7)+3=17$ .
2	$E(X)=8$ $\operatorname{Var}(X)=4.8$ $1-p=0.6$ $p=0.4$ $n=20$ $n$ is a positive integer and $0<p<1$	6
(6 marks)		6
Notes
$5-2E(X)=-11$ gives $E(X)=np=8$ . Also $\operatorname{Var}(Y)=4\operatorname{Var}(X)$ , so $\operatorname{Var}(X)=19.2/4=4.8$ . Dividing $np(1-p)=4.8$ by $np=8$ gives $1-p=0.6$ , hence $p=0.4$ and $n=8/0.4=20$ . These satisfy the binomial restrictions.
3	$\operatorname{Var}(X)=88.2-9^2=7.2$ $np=9$ and $np(1-p)=7.2$ $1-p=0.8$ $p=0.2$ $n=45$ $P(X=0)=(0.8)^{45}=0.00004356$ to $4$ significant figures	6
(6 marks)		6
Notes
$\operatorname{Var}(X)=E(X^2)-[E(X)]^2=88.2-81=7.2$ . Hence $np=9$ and $np(1-p)=7.2$ . Dividing gives $1-p=7.2/9=0.8$ , so $p=0.2$ and $n=9/0.2=45$ . Therefore $P(X=0)=(1-p)^n=(0.8)^{45}=0.0000435561\ldots$ , which is $0.00004356$ to $4$ significant figures.

Tier 3 · Hard

Mark scheme for FS1-2.2 Tier 3 · Hard
Question	Scheme	Marks
1	$n=72$ $p=0.25$ $P(X=18)=\binom{72}{18}(0.25)^{18}(0.75)^{54}=0.1080$ to $4$ significant figures	6
(6 marks)		6
Notes
Use $np=18$ and $np(1-p)=13.5$ . Dividing the second equation by the first gives $1-p=13.5/18=0.75$ , so $p=0.25$ . Then $n=18/0.25=72$ . Therefore $P(X=18)=\binom{72}{18}(0.25)^{18}(0.75)^{54}=0.1080351845\ldots=0.1080$ to $4$ significant figures.
2	$E(X)=12$ $\operatorname{Var}(X)=3$ $np=12$ and $np(1-p)=3$ $1-p=0.25$ $p=0.75$ $n=16$ $P(X\geq15)=16(0.75)^{15}(0.25)+(0.75)^{16}=0.0635$ to $3$ significant figures	7
(7 marks)		7
Notes
$3E(X)-2=34$ gives $E(X)=np=12$ , while $9\operatorname{Var}(X)=27$ gives $np(1-p)=3$ . Dividing the variance equation by the mean equation gives $1-p=0.25$ , so $p=0.75$ and $n=16$ . Therefore $P(X\geq15)=P(X=15)+P(X=16)=16(0.75)^{15}(0.25)+(0.75)^{16}=0.0634764398\ldots=0.0635$ to $3$ significant figures.
3	$50p(1-p)=8$ $p^2-p+0.16=0$ , so $p=0.2$ or $p=0.8$ $E(X)>25$ selects $p=0.8$ $E(X)=50(0.8)=40$ $\operatorname{sd}(X)=\sqrt8=2\sqrt2$ $P(38\leq X\leq40)=\displaystyle\sum_{x=38}^{40}\binom{50}{x}(0.8)^x(0.2)^{50-x}$ $P(38\leq X\leq40)=0.3702$ to $4$ significant figures	7
(7 marks)		7
Notes
The variance condition gives $50p(1-p)=8$ , or $p^2-p+0.16=0$ , so $p=0.2$ or $p=0.8$ . Since $E(X)=50p>25$ , the required value is $p=0.8$ . Therefore $E(X)=40$ and the standard deviation is $\sqrt8=2\sqrt2$ . Summing the three binomial probabilities gives $P(38\leq X\leq40)=\sum_{x=38}^{40}\binom{50}{x}(0.8)^x(0.2)^{50-x}=0.3702025932\ldots$ , which is $0.3702$ to $4$ significant figures.
4	$n(0.35)(0.65)=9.1$ $n=40$ $E(X)=40(0.35)=14$ $E(W)=4E(X)-7=49$ $\operatorname{Var}(W)=16\operatorname{Var}(X)$ $\operatorname{Var}(W)=145.6$ $41\leq W\leq57$ is equivalent to $12\leq X\leq16$ $P(41\leq W\leq57)=\displaystyle\sum_{x=12}^{16}\binom{40}{x}(0.35)^x(0.65)^{40-x}=0.5925$	8
(8 marks)		8
Notes
The binomial variance is $n(0.35)(0.65)=9.1$ , so $n=40$ . Hence $E(X)=40(0.35)=14$ . For $W=4X-7$ , linearity gives $E(W)=4(14)-7=49$ , while $\operatorname{Var}(W)=4^2(9.1)=145.6$ . The inequalities $41\leq4X-7\leq57$ give $12\leq X\leq16$ . Therefore the required binomial sum is $\sum_{x=12}^{16}\binom{40}{x}(0.35)^x(0.65)^{40-x}=0.5924796652\ldots=0.5925$ to $4$ decimal places.
5	$\dfrac{P(X=n)}{P(X=n-1)}=\dfrac{p}{n(1-p)}$ $\dfrac{p}{n(1-p)}=0.1$ $np=9$ $n^2-9n-90=0$ $n=15$ $p=0.6$ $\operatorname{Var}(X)=3.6$ $P(X\geq14)=15(0.6)^{14}(0.4)+(0.6)^{15}=0.005172$	8
(8 marks)		8
Notes
The ratio of the last two binomial probabilities is $p/[n(1-p)]$ , so $p/[n(1-p)]=0.1$ . The mean condition gives $p=9/n$ . Substitution gives $9/[n(n-9)]=0.1$ , hence $n^2-9n-90=0$ . Its roots are $15$ and $-6$ , so $n=15$ and $p=0.6$ . Therefore $\operatorname{Var}(X)=np(1-p)=3.6$ and $P(X\geq n-1)=P(X=14)+P(X=15)=15(0.6)^{14}(0.4)+(0.6)^{15}=0.0051720348\ldots=0.005172$ .

FS1-2.3 · The use of the Poisson distribution as an approximation to the binomial distribution.

Tier 1 · Easy

Mark scheme for FS1-2.3 Tier 1 · Easy
Question	Scheme	Marks
1	$X\mathrel{\dot\sim}\operatorname{Po}(3.6)$ $P(X=0)\approx0.02732$	3
(3 marks)		3
Notes
Here $n$ is large, $p$ is small and $np=3.6$ , so use $Y\sim\operatorname{Po}(3.6)$ . Then $P(X=0)\approx P(Y=0)=e^{-3.6}=0.0273237\ldots$ .
2	Since $n$ is large and $p$ is small, use $Y\sim\operatorname{Po}(3)$ $P(Y\leq4)=0.815263\ldots<0.9$ but $P(Y\leq5)=0.916082\ldots\geq0.9$ $k=5$	3
(3 marks)		3
Notes
Here $np=3$ , with large $n$ and small $p$ , so use $Y\sim\operatorname{Po}(3)$ . The adjacent cumulative probabilities are $0.8152632445\ldots$ at $4$ and $0.9160820580\ldots$ at $5$ , proving that $k=5$ is the smallest possible integer.

Tier 2 · Standard

Mark scheme for FS1-2.3 Tier 2 · Standard
Question	Scheme	Marks
1	$n=400$ is large and $p=0.01$ is small, so a Poisson approximation is suitable Use $Y\sim\operatorname{Po}(4)$ because $np=400(0.01)=4$ $P(X\geq6)\approx P(Y\geq6)=1-P(Y\leq5)$ $P(X\geq6)\approx0.2149$ to $4$ significant figures	4
(4 marks)		4
Notes
Here $n=400$ is large, $p=0.01$ is small and $np=400(0.01)=4$ , so $X\mathrel{\dot\sim}Y$ where $Y\sim\operatorname{Po}(4)$ . Thus $P(X\geq6)\approx1-P(Y\leq5)=0.2148696129\ldots=0.2149$ to $4$ significant figures.
2	Use $Y\sim\operatorname{Po}(4)$ because $np=4$ $P(3\leq Y\leq6)=\sum_{y=3}^{6}e^{-4}\dfrac{4^y}{y!}$ The Poisson approximation is $0.6512$ to $4$ decimal places $n=2000$ is large and $p=0.002$ is small, so the approximation is valid At $p=0.2$ , the success probability is not small, so a Poisson approximation would not be valid	5
(5 marks)		5
Notes
Here $np=2000(0.002)=4$ , so use $Y\sim\operatorname{Po}(4)$ . Then $P(3\leq X\leq6)\approx P(3\leq Y\leq6)=\sum_{y=3}^{6}e^{-4}4^y/y!=0.651222716043882\ldots$ , giving $0.6512$ to $4$ decimal places. The conditions are met because $n$ is large and $p$ is small. If $p=0.2$ , the success probability is not small, so a Poisson approximation would not be valid.
3	$E(X)=0.003n=5.4$ $n=1800$ $n$ is large and $p=0.003$ is small, so a Poisson approximation is suitable Use $Y\sim\operatorname{Po}(5.4)$ $P(4\leq Y\leq7)=0.6084$ to $4$ decimal places	5
(5 marks)		5
Notes
The binomial mean is $np$ , so $0.003n=5.4$ and $n=1800$ . This is large and the flaw probability is small, so use $Y\sim\operatorname{Po}(5.4)$ . Then $P(4\leq X\leq7)\approx P(4\leq Y\leq7)=\sum_{y=4}^{7}e^{-5.4}5.4^y/y!=0.6083676684\ldots=0.6084$ to $4$ decimal places.

Tier 3 · Hard

Mark scheme for FS1-2.3 Tier 3 · Hard
Question	Scheme	Marks
1	Exact probability $=0.42150$ Poisson approximation $=0.42319$ Percentage error $=0.402\%$	6
(6 marks)		6
Notes
Exactly, $P(X\leq2)=\sum_{r=0}^{2}\binom{200}{r}(0.015)^r(0.985)^{200-r}=0.4214963\ldots$ . Since $np=3$ , use $Y\sim\operatorname{Po}(3)$ , giving $P(Y\leq2)=0.4231901\ldots$ . The relative percentage error is $100(0.4231901-0.4214963)/0.4214963=0.40184\ldots\%$ .
2	$X\sim\operatorname{Bin}(n,0.004)$ is approximated by $Y\sim\operatorname{Po}(0.004n)$ $P(Y\leq1)=e^{-0.004n}(1+0.004n)$ The continuous boundary parameter is $3.8897201699\ldots$ $n=972$ gives probability $0.1001369251\ldots>0.1$ $n=973$ gives probability $0.0998187955\ldots<0.1$ The smallest sample size is $973$	6
(6 marks)		6
Notes
For large $n$ and small $p=0.004$ , use $Y\sim\operatorname{Po}(0.004n)$ . Solving $e^{-\lambda}(1+\lambda)=0.1$ gives $\lambda=3.8897201699\ldots$ . The adjacent integers give probabilities $0.1001369251\ldots$ for $n=972$ and $0.0998187955\ldots$ for $n=973$ , so $n=973$ is minimal.
3	For $Y\sim\operatorname{Po}(\lambda)$ , $\dfrac{P(Y=3)}{P(Y=2)}=\lambda/3$ $\lambda/3=1.4$ , so $\lambda=4.2$ $500p=4.2$ , so $p=0.0084$ Use $Y\sim\operatorname{Po}(4.2)$ $P(Y\geq6)=0.2469$ to $4$ decimal places The approximation is reasonable because $n=500$ is large and the estimated value of $p$ is small	6
(6 marks)		6
Notes
For $Y\sim\operatorname{Po}(\lambda)$ , the ratio of the two point probabilities is $[e^{-\lambda}\lambda^3/3!]/[e^{-\lambda}\lambda^2/2!]=\lambda/3$ . Hence $\lambda/3=1.4$ , giving $\lambda=4.2$ . Since the approximating parameter is $500p$ , $p=4.2/500=0.0084$ . Therefore $P(Y\geq6)=1-P(Y\leq5)=0.2468571113\ldots=0.2469$ . The estimated probability is small while $500$ is large, so the approximation is reasonable.
4	$X$ is approximated by $X'\sim\operatorname{Po}(2.4)$ $Y$ is approximated by $Y'\sim\operatorname{Po}(3.5)$ Split the event into $X'=0,Y'\geq5$ and $X'=1,Y'\geq4$ $P(X'=0)=e^{-2.4}$ and $P(X'=1)=2.4e^{-2.4}$ $P(Y'\geq5)=0.2745550\ldots$ $P(Y'\geq4)=0.4633673\ldots$ $P(X\leq1\text{ and }X+Y\geq5)\approx e^{-2.4}P(Y'\geq5)+2.4e^{-2.4}P(Y'\geq4)=0.1258$	7
(7 marks)		7
Notes
The two small flaw probabilities and large batch sizes give the independent approximations $X'\sim\operatorname{Po}(800\times0.003)=\operatorname{Po}(2.4)$ and $Y'\sim\operatorname{Po}(500\times0.007)=\operatorname{Po}(3.5)$ . If $X'\leq1$ and $X'+Y'\geq5$ , either $X'=0,Y'\geq5$ or $X'=1,Y'\geq4$ . Independence therefore gives $e^{-2.4}(0.2745550467\ldots)+2.4e^{-2.4}(0.4633673321\ldots)=0.1257928383\ldots$ , which is $0.1258$ to $4$ decimal places.
5	Each batch count is approximated by $\operatorname{Po}(0.002n)$ The probability that one batch has no flaws is $q=e^{-0.002n}$ The probability that at least one of three batches has no flaws is $1-(1-q)^3$ $1-(1-q)^3\geq0.5$ gives $q\geq1-2^{-1/3}$ $n\leq\dfrac{-\ln(1-2^{-1/3})}{0.002}=789.213\ldots$ For $n=789$ , the probability is $0.500166\ldots$ For $n=790$ , the probability is $0.499387\ldots$ The largest integer is $n=789$	8
(8 marks)		8
Notes
Here $p=0.002$ is small and $n$ is large, so $\operatorname{Po}(0.002n)$ is appropriate for each batch count. Its zero-flaw probability is $q=e^{-0.002n}$ . Independence between batches gives probability $1-(1-q)^3$ that at least one has zero flaws. The condition is equivalent to $q\geq1-2^{-1/3}$ and hence $n\leq-\ln(1-2^{-1/3})/0.002=789.2132043\ldots$ . The adjacent checks give $0.5001662658\ldots$ at $n=789$ and $0.4993867176\ldots$ at $n=790$ , so $789$ is the largest integer.