Geometric and negative binomial distributions

FS1-3 Geometric and negative binomial distributions — revision question pack

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

Checked against Edexcel 9FM0 section FS1-3. 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.

How this checking works

FS1-3.1 · Geometric and negative binomial distributions.

Explanation

For independent Bernoulli trials with constant success probability $p$ , a geometric variable $X$ counts the trial of the first success: $P(X=x)=p(1-p)^{x-1}$ for $x=1,2,\ldots$ .
A negative binomial variable $Y$ counts the trial of the $r$ th success: $P(Y=y)=\binom{y-1}{r-1}p^r(1-p)^{y-r}$ for $y=r,r+1,\ldots$ .
The last trial is fixed as a success, while $r-1$ successes are placed among the preceding $y-1$ trials.
Cumulative events can sometimes be recast as binomial counts.
Examiners expect the support convention, the final-success condition and the independence/constant- $p$ modelling assumptions to be clear.

Worked example

Independent trials have success probability $0.3$ . Find the probability that the fourth success occurs on trial $7$ .

1.Trial $7$ must be a success.
2.Exactly $3$ successes must occur among the first $6$ trials, in $\binom63$ arrangements.
3. $P(Y=7)=\binom63(0.3)^4(0.7)^3=0.05557$ to $4$ significant figures.

Answer: The probability is $0.05557$ .

Common mistakes

Don't use $\binom74$ and allow the final trial not to be a success.
Don't use exponent $y-r+1$ for failures instead of $y-r$ .
Don't start a geometric variable at $0$ despite the FS₁ convention counting the success trial.

Exam tip

Write 'final trial is a success' before choosing the negative-binomial combination.

Tier 1 · Easy

1.
Independent trials have success probability $0.3$ . Find the probability that the first success occurs on trial $4$ .
(2)
(Total for Question 1 is 2 marks)
2.
Independent trials succeed with probability $p$ . The random variable $Y$ is the trial on which the third success occurs. Show that $P(Y=y)=\binom{y-1}{2}p^3(1-p)^{y-3}$ for $y\geq3$ . Given $p=0.25$ , find $P(Y=6)$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Independent trials have success probability $0.4$ . Find the probability that the third success occurs on trial $5$ , and find the probability that the third success occurs after trial $6$ .
(4)
(Total for Question 1 is 4 marks)
2.
A player succeeds independently on each attempt with probability $0.4$ . Let $X$ be the attempt on which the fifth success occurs. Find the probability that $X=8$ and exactly two of the first four attempts are successes.
(5)
(Total for Question 2 is 5 marks)
3.
Independent trials succeed with probability $0.45$ . Let $U$ and $V$ be the trials on which the third and fifth successes occur respectively. Find $P(V-U=3)$ and $P(V-U\leq3)$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Each route attempt succeeds independently with probability $0.35$ . Calculate the probability that the second successful route is completed by the fifth attempt.
(5)
(Total for Question 1 is 5 marks)
2.
Independent trials succeed with probability $0.4$ . Sampling stops as soon as either two successes or three failures have occurred. Find the probability that the two-success condition ends the sampling and find the expected number of trials used.
(7)
(Total for Question 2 is 7 marks)
3.
Independent trials have success probability $p$ , where $0<p<1$ . The random variable $Y$ is the trial on which the fourth success occurs. It is known that $P(Y=10)=P(Y=11)$ . Determine $p$ , find all the most likely values of $Y$ , and calculate $P(Y=10)$ and $P(Y\leq6)$ .
(8)
(Total for Question 3 is 8 marks)
4.
A controller first chooses target A with probability $2/5$ or target B with probability $3/5$ . It then observes independent trials with success probability $2/7$ , stopping at the second success for target A or at the fourth success for target B. Let $T$ be the stopping trial. Find $P(T=5)$ , $P(\text{target A}\mid T=5)$ and $P(T=4)$ .
(7)
(Total for Question 4 is 7 marks)
5.
Independent trials have success probability $p$ , where $0<p<1$ . The random variable $Y$ is the trial on which the fifth success occurs. Given that $Y=12$ , find the conditional probability that the third success occurred by trial $6$ , the conditional probability that the first three trials all failed, and the conditional probability that exactly one of the first four trials succeeded.
(8)
(Total for Question 5 is 8 marks)

FS1-3.2 · Mean and variance of a geometric distribution with parameter p.

Explanation

For a geometric variable counting the trial of the first success, $E(X)=1/p$ and $\operatorname{Var}(X)=(1-p)/p^2$ . Its standard deviation is $\sqrt{1-p}/p$ .
These formulas use support $1,2,\ldots$ , not the alternative convention that counts failures before success.
Given the mean or variance, solve for a probability satisfying $0<p\leq1$ and check that the result is consistent.
The memoryless property explains why, after any run of failures, the remaining waiting time has the same geometric model.
Examiners expect the convention to match the question, the standard deviation to be positive, and probability events such as $X\geq k$ to count the required initial failures correctly.

Worked example

A geometric random variable has variance $20$ . Find $p$ , its mean and $P(X>3)$ .

1. $\dfrac{1-p}{p^2}=20$ , so $20p^2+p-1=0$ .
2.Since $0<p\leq1$ , the only possible root is $p=0.2$ ; the other root is negative.
3. $E(X)=1/p=5$ and $P(X>3)=(1-p)^3=0.8^3=0.512$ .

Answer: $p=0.2$ , mean $5$ , and $P(X>3)=0.512$ .

Common mistakes

Don't use $(1-p)/p$ rather than $(1-p)/p^2$ for the variance.
Don't keep a negative quadratic root for the probability parameter.
Don't interpret $X>3$ as four failures rather than failure on the first three trials.

Exam tip

After solving for $p$ , reject any root outside $0<p\leq1$ explicitly.

Tier 1 · Easy

1.
The random variable $X$ is geometric with parameter $p=0.2$ and counts the trial of the first success. Find its mean and variance.
(2)
(Total for Question 1 is 2 marks)
2.
$X$ is geometric with $p=0.4$ . Find $E(X)$ , $\operatorname{Var}(X)$ and $P(X\geq E(X))$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
A geometric random variable has mean $8$ . Find $p$ , its variance and its standard deviation.
(4)
(Total for Question 1 is 4 marks)
2.
The geometric variable $X$ counts the trial of the first success. Given that $P(X\geq7\mid X\geq3)=81/256$ , determine $p$ , the mean and the variance.
(5)
(Total for Question 2 is 5 marks)
3.
A geometric random variable $X$ with $0<p<1$ counts the trial of the first success. It is known that $P(X=6)=\dfrac58P(X=5)$ . Determine $p$ , then find $E(X)$ , $\operatorname{Var}(X)$ and $P(X\leq4)$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
For a geometric random variable $X$ , the variance is twice the mean. Determine $p$ , then find $P(X\geq4)$ , $E(X)$ and $\operatorname{Var}(X)$ .
(6)
(Total for Question 1 is 6 marks)
2.
A geometric random variable $X$ counts the trial of the first success. Find the largest value of $p$ for which $P(X\geq4)\geq0.1$ , then find the mean and the variance, giving all numerical answers to $3$ significant figures.
(6)
(Total for Question 2 is 6 marks)
3.
A geometric random variable $X$ counts the trial of the first success, where $0<p<1$ . Given that $E(X)+\sqrt{\operatorname{Var}(X)}=6$ , determine $p$ , $E(X)$ and $\operatorname{Var}(X)$ . Hence find $P(X\leq3)$ exactly.
(7)
(Total for Question 3 is 7 marks)
4.
A geometric random variable $X$ counts the trial of the first success. Its standard deviation divided by its mean is $3/4$ . Determine $p$ , $E(X)$ and $\operatorname{Var}(X)$ . Find the least integer $k$ for which $P(X\leq k)\geq0.95$ .
(8)
(Total for Question 4 is 8 marks)
5.
Independent geometric random variables $X$ and $Y$ count the trials of their respective first successes. It is known that $E(X)=2$ and $\operatorname{Var}(Y)=30$ . Determine the two success probabilities, then find $P(X<Y)$ , $P(X=Y)$ and $P(X>Y)$ exactly.
(8)
(Total for Question 5 is 8 marks)

FS1-3.3 · Mean and variance of negative binomial distribution.

Explanation

If $X$ counts the trial on which the $r$ th success occurs, then $E(X)=r/p$ and $\operatorname{Var}(X)=r(1-p)/p^2$ .
The ratio $\operatorname{Var}(X)/E(X)=(1-p)/p$ is efficient when both moments are known: it determines $p$ , then the mean determines the positive integer $r$ .
These are formulas for the success-trial number, not for the number of failures before $r$ successes.
Probabilities still use $\binom{x-1}{r-1}p^r(1-p)^{x-r}$ .
Examiners expect any recovered $r$ to be a positive integer, $p$ to lie in $(0,1]$ , and cumulative boundaries to respect the minimum value $X=r$ .

Worked example

A negative binomial variable has mean $12$ and variance $24$ . Find $p$ and $r$ .

1. $\dfrac{\operatorname{Var}(X)}{E(X)}=2=\dfrac{1-p}{p}$ .
2.Thus $2p=1-p$ , so $p=\dfrac13$ .
3. $r/p=12$ , hence $r=4$ .

Answer: $p=\dfrac13$ and $r=4$ .

Common mistakes

Don't use the mean number of failures $r(1-p)/p$ instead of the mean trial number $r/p$ .
Don't cancel the ratio of variance to mean to $(1-p)$ and lose the factor $1/p$ .
Don't accept a non-integer value of $r$ for a count of successes.

Exam tip

When both moments are given, take variance divided by mean before solving for $r$ .

Tier 1 · Easy

1.
The random variable $X$ counts the trial on which the fourth success occurs, with success probability $0.25$ . Find $E(X)$ and $\operatorname{Var}(X)$ .
(2)
(Total for Question 1 is 2 marks)
2.
A negative binomial variable $X$ counts the trial of the $r$ th success. Given $p=0.2$ and $\operatorname{Var}(X)=100$ , find $r$ , $E(X)$ and $P(X=7)$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
A negative binomial random variable has mean $15$ and variance $30$ . Find $p$ and $r$ , then calculate $P(X=7)$ .
(6)
(Total for Question 1 is 6 marks)
2.
A negative binomial variable $X$ counts the trial of the $r$ th success. The parameters satisfy $p=2/r$ and $\operatorname{Var}(X)=8$ . Find $r$ and $p$ , then find $E(X)$ and $P(X=6)$ .
(7)
(Total for Question 2 is 7 marks)
3.
The random variable $X$ counts the trial on which the $r$ th success occurs. It is known that $P(X\leq2)=0$ , $P(X=3)>0$ , and the standard deviation of $X$ is one third of its mean. Determine $r$ and $p$ , find $E(X)$ , and show that $P(X=3)=P(X=4)$ .
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Successive trials are independent with success probability $0.4$ . Let $X$ be the trial on which the sixth success occurs. Find the probability that $X\leq8$ . Conditional on $X\leq8$ , find the probability that $X=8$ and the expected value of $X$ .
(7)
(Total for Question 1 is 7 marks)
2.
A negative binomial variable $X$ counts the trial of the $r$ th success. It satisfies $\operatorname{Var}(X)=3E(X)$ and $P(X=r)=1/1024$ . Determine $p$ and $r$ , verify that $r$ is a positive integer, and find $E(X)$ , $\operatorname{Var}(X)$ and $P(X\leq7)$ .
(8)
(Total for Question 2 is 8 marks)
3.
The random variable $X$ counts the trial on which the $r$ th success occurs. Let $F=X-r$ be the number of failures before the $r$ th success. Given that $E(F)=7$ and $\operatorname{Var}(F)=14$ , determine $r$ and $p$ . Show that $E(F)=\operatorname{Var}(F)$ is impossible for $0<p<1$ , and find $P(F\leq3)$ .
(8)
(Total for Question 3 is 8 marks)

Answer key

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

FS1-3.1 · Geometric and negative binomial distributions.

Tier 1 · Easy

Mark scheme for FS1-3.1 Tier 1 · Easy
Question	Scheme	Marks
1	$0.1029$	2
(2 marks)		2
Notes
The first three trials must fail and trial $4$ must succeed, so the probability is $(0.7)^3(0.3)=0.1029$ .
2	Exactly two of the first $y-1$ trials must succeed Trial $y$ must be a success $P(Y=y)=\binom{y-1}{2}p^3(1-p)^{y-3}$ for $y\geq3$ $P(Y=6)=\binom52(0.25)^3(0.75)^3=\dfrac{135}{2048}=0.065918$ to $6$ decimal places	4
(4 marks)		4
Notes
For the third success to occur on trial $y$ , exactly two of the first $y-1$ trials must succeed and trial $y$ must succeed. This gives $\binom{y-1}{2}p^2(1-p)^{y-3}p$ . With $p=0.25$ and $y=6$ , the probability is $\binom52(0.25)^3(0.75)^3=135/2048=0.06591796875$ .

Tier 2 · Standard

Mark scheme for FS1-3.1 Tier 2 · Standard
Question	Scheme	Marks
1	$P(Y=5)=0.13824$ $P(Y>6)=0.54432$	4
(4 marks)		4
Notes
Exactly two of the first four trials must be successes, followed by a success. Hence $P(Y=5)=\binom{4}{2}(0.4)^3(0.6)^2=0.13824$ . The third success occurs after trial $6$ exactly when at most two of the first six trials succeed, so $P(Y>6)=\sum_{k=0}^{2}\binom6k(0.4)^k(0.6)^{6-k}=0.54432$ .
2	Exactly two of attempts $1$ -- $4$ succeed in $\binom42$ ways Exactly two of attempts $5$ -- $7$ must succeed, in $\binom32$ ways Attempt $8$ must be the fifth success $P(X=8\text{ and exactly two successes in attempts 1--4})=\binom42\cdot\binom32(0.4)^5(0.6)^3$ The probability is $18(0.4)^5(0.6)^3=0.0398$ to $3$ significant figures	5
(5 marks)		5
Notes
For the fifth success to occur on attempt $8$ , the first seven attempts must contain four successes and attempt $8$ must succeed. The contextual constraint allocates two successes to attempts $1$ -- $4$ , so attempts $5$ -- $7$ must contain the other two. Hence the probability is $\binom42\binom32(0.4)^4(0.6)^3(0.4)=18(0.4)^5(0.6)^3=0.03981312=0.0398$ to $3$ significant figures.
3	After trial $U$ , the gap $V-U$ counts trials until two further successes $P(V-U=3)=\binom21(0.45)^2(0.55)$ $P(V-U=3)=0.22275$ $P(V-U=2)=(0.45)^2=0.2025$ $P(V-U\leq3)=0.2025+0.22275=0.42525$	5
(5 marks)		5
Notes
Once the third success occurs, independence means the additional wait for two successes starts afresh. For a gap of $3$ , exactly one of the next two trials succeeds and the third trial succeeds, giving $\binom21(0.45)^2(0.55)=0.22275$ . The smallest possible gap is $2$ , with probability $(0.45)^2=0.2025$ . Direct summation therefore gives $P(V-U\leq3)=P(V-U=2)+P(V-U=3)=0.2025+0.22275=0.42525$ .

Tier 3 · Hard

Mark scheme for FS1-3.1 Tier 3 · Hard
Question	Scheme	Marks
1	$0.571585$	5
(5 marks)		5
Notes
The second success occurs by attempt $5$ exactly when there are at least two successes among the first five attempts. Its complement has zero or one success, so the probability is $1-(0.65)^5-\binom51(0.35)(0.65)^4=0.571585$ .
2	A success win on trial $2$ has probability $0.4^2$ A success win on trial $3$ has probability $2(0.4)^2(0.6)$ A success win on trial $4$ has probability $3(0.4)^2(0.6)^2$ The success condition wins with probability $0.5248$ $P(T=2)=0.16$ $P(T=3)=2(0.4)^2(0.6)+0.6^3=0.408$ $E(T)=2(0.16)+3(0.408)+4(0.432)=3.272$	7
(7 marks)		7
Notes
For two successes to win on trials $2,3,4$ , the exact chains have probabilities $0.4^2$ , $2(0.4)^2(0.6)$ and $3(0.4)^2(0.6)^2$ . Their sum is $0.16+0.192+0.1728=0.5248$ . The stopping-time probabilities are $P(T=2)=0.16$ , $P(T=3)=2(0.4)^2(0.6)+0.6^3=0.408$ , and $P(T=4)=1-0.16-0.408=0.432$ . Hence $E(T)=2(0.16)+3(0.408)+4(0.432)=3.272$ .
3	$P(Y=10)=\binom93p^4(1-p)^6$ $P(Y=11)=\binom{10}3p^4(1-p)^7$ $\binom93=\binom{10}3(1-p)$ , so $1-p=0.7$ $p=0.3$ $\dfrac{P(Y=y+1)}{P(Y=y)}=\dfrac{(1-p)y}{y-3}$ changes through $1$ at $y=10$ The two most likely values are $Y=10$ and $Y=11$ $P(Y=10)=\binom93(0.3)^4(0.7)^6=0.08005$ to $4$ significant figures $P(Y\leq6)=\displaystyle\sum_{y=4}^{6}\binom{y-1}3(0.3)^4(0.7)^{y-4}=0.07047$	8
(8 marks)		8
Notes
Equality of the two probabilities gives $\binom93p^4(1-p)^6=\binom{10}3p^4(1-p)^7$ . Cancelling positive common factors yields $84=120(1-p)$ , so $p=0.3$ . For the fourth-success distribution, the consecutive-probability ratio is $(1-p)y/(y-3)$ . With $p=0.3$ this ratio is greater than $1$ before $y=10$ , equal to $1$ at $y=10$ and below $1$ afterwards, so $10$ and $11$ are equal modes. Substitution gives $P(Y=10)=\binom93(0.3)^4(0.7)^6=0.0800483796\ldots$ . Direct summation over the support values $4,5,6$ gives $P(Y\leq6)=\sum_{y=4}^{6}\binom{y-1}3(0.3)^4(0.7)^{y-4}=0.07047$ .
4	$P(T=5\mid\text{target A})=\binom41(2/7)^2(5/7)^3=2000/16807$ $P(T=5\mid\text{target B})=\binom43(2/7)^4(5/7)=320/16807$ $P(T=5)=\dfrac25\left(\dfrac{2000}{16807}\right)+\dfrac35\left(\dfrac{320}{16807}\right)=\dfrac{992}{16807}$ $P(\text{target A}\cap T=5)=\dfrac25\left(\dfrac{2000}{16807}\right)=\dfrac{800}{16807}$ $P(\text{target A}\mid T=5)=\dfrac{P(\text{target A}\cap T=5)}{P(T=5)}=\dfrac{25}{31}$ $P(T=4\mid\text{target A})=\binom31(2/7)^2(5/7)^2=300/2401$ and $P(T=4\mid\text{target B})=(2/7)^4=16/2401$ $P(T=4)=\dfrac25\left(\dfrac{300}{2401}\right)+\dfrac35\left(\dfrac{16}{2401}\right)=\dfrac{648}{12005}$	7
(7 marks)		7
Notes
For target A, trial $5$ must be a success and exactly one of the first four trials succeeds, giving $\binom41(2/7)^2(5/7)^3=2000/16807$ . For target B, exactly three of the first four trials succeed and trial $5$ succeeds, giving $\binom43(2/7)^4(5/7)=320/16807$ . Total probability gives $P(T=5)=(2/5)(2000/16807)+(3/5)(320/16807)=992/16807$ . Therefore $P(\text{target A}\mid T=5)=P(\text{target A}\cap T=5)/P(T=5)=[(2/5)(2000/16807)]/(992/16807)=25/31$ . For $T=4$ , target A requires exactly one success in the first three trials and a success on trial $4$ , while target B requires four consecutive successes. Hence $P(T=4)=(2/5)(300/2401)+(3/5)(16/2401)=648/12005$ .
5	Given $Y=12$ , exactly four of the first $11$ trials succeeded The $\binom{11}{4}$ possible success-position sets are conditionally equally likely $P(\text{third success by trial }6\mid Y=12)=\dfrac{\binom63\binom51+\binom64\binom50}{\binom{11}{4}}$ $P(\text{third success by trial }6\mid Y=12)=\dfrac{23}{66}$ $P(\text{first three all fail}\mid Y=12)=\dfrac{\binom84}{\binom{11}{4}}$ $P(\text{first three all fail}\mid Y=12)=\dfrac7{33}$ $P(\text{exactly one success in first four}\mid Y=12)=\dfrac{\binom41\binom73}{\binom{11}{4}}$ $P(\text{exactly one success in first four}\mid Y=12)=\dfrac{14}{33}$	8
(8 marks)		8
Notes
The condition $Y=12$ means that exactly four successes occupy the first $11$ positions and trial $12$ is a success. Conditional on this total, every choice of four positions has the same factor $p^4(1-p)^7$ , so the $\binom{11}{4}$ sets are equally likely. The third success is by trial $6$ when the first six contain three or four successes, giving numerator $\binom63\binom51+\binom64\binom50=115$ and probability $115/330=23/66$ . If the first three all fail, all four successes lie among the next eight positions, giving $\binom84/\binom{11}{4}=7/33$ . Exactly one success in the first four gives $\binom41\binom73/\binom{11}{4}=14/33$ .

FS1-3.2 · Mean and variance of a geometric distribution with parameter p.

Tier 1 · Easy

Mark scheme for FS1-3.2 Tier 1 · Easy
Question	Scheme	Marks
1	$E(X)=5$ $\operatorname{Var}(X)=20$	2
(2 marks)		2
Notes
$E(X)=1/p=1/0.2=5$ and $\operatorname{Var}(X)=(1-p)/p^2=0.8/0.2^2=20$ .
2	$E(X)=1/p=2.5$ $\operatorname{Var}(X)=(1-p)/p^2=3.75$ Since $X$ is integer-valued, $X\geq E(X)$ means $X\geq3$ $P(X\geq E(X))=P(X\geq3)=0.6^2=0.36$	4
(4 marks)		4
Notes
For $p=0.4$ , $E(X)=1/p=2.5$ and $\operatorname{Var}(X)=(1-p)/p^2=0.6/0.16=3.75$ . Because $X$ takes integer values, $X\geq2.5$ is the event $X\geq3$ , which requires two initial failures. Hence $P(X\geq E(X))=0.6^2=0.36$ .

Tier 2 · Standard

Mark scheme for FS1-3.2 Tier 2 · Standard
Question	Scheme	Marks
1	$p=0.125$ Variance $=56$ Standard deviation $=7.48$ to $3$ significant figures	4
(4 marks)		4
Notes
$1/p=8$ , so $p=1/8=0.125$ . Then $\operatorname{Var}(X)=(1-1/8)/(1/8)^2=56$ , giving standard deviation $\sqrt{56}=7.4833\ldots$ .
2	$P(X\geq7\mid X\geq3)=(1-p)^4$ $(1-p)^4=(3/4)^4$ $p=1/4$ $E(X)=4$ $\operatorname{Var}(X)=12$	5
(5 marks)		5
Notes
Conditional on the first two trials having failed, $X\geq7$ requires four further failures. Hence $P(X\geq7\mid X\geq3)=(1-p)^4=81/256=(3/4)^4$ . Since $0<p\leq1$ , the only possible solution is $p=1/4$ . Thus $E(X)=1/p=4$ and $\operatorname{Var}(X)=(1-p)/p^2=(3/4)/(1/16)=12$ .
3	$\dfrac{P(X=6)}{P(X=5)}=1-p=\dfrac58$ $p=3/8$ $E(X)=8/3$ $\operatorname{Var}(X)=40/9$ $P(X\leq4)=1-(5/8)^4=3471/4096$	5
(5 marks)		5
Notes
Since $P(X=x)=p(1-p)^{x-1}$ , the ratio of adjacent probabilities is $1-p$ . Thus $1-p=5/8$ and $p=3/8$ . The moments are $E(X)=1/p=8/3$ and $\operatorname{Var}(X)=(1-p)/p^2=(5/8)/(9/64)=40/9$ . Finally, $P(X\leq4)=1-P(X>4)=1-(5/8)^4=3471/4096$ .

Tier 3 · Hard

Mark scheme for FS1-3.2 Tier 3 · Hard
Question	Scheme	Marks
1	$p=\frac13$ $P(X\geq4)=\frac{8}{27}$ $E(X)=3$ $\operatorname{Var}(X)=6$	6
(6 marks)		6
Notes
Set $(1-p)/p^2=2/p$ . Multiplying by $p^2$ gives $1-p=2p$ , so $p=1/3$ . The event $X\geq4$ means the first three trials fail, so $P(X\geq4)=(2/3)^3=8/27$ . The mean is $1/p=3$ and the variance is $(1-p)/p^2=6$ .
2	$P(X\geq4)=(1-p)^3$ The boundary is $(1-p)^3=0.1$ $p=1-0.1^{1/3}=0.536$ to $3$ significant figures This is the largest value because $(1-p)^3$ decreases as $p$ increases $E(X)=1/p=1.87$ to $3$ significant figures $\operatorname{Var}(X)=(1-p)/p^2=1.62$ to $3$ significant figures	6
(6 marks)		6
Notes
The event $X\geq4$ means three initial failures, so $(1-p)^3\geq0.1$ . Since this probability decreases with $p$ , the largest value occurs at equality: $p=1-0.1^{1/3}=0.5358411166\ldots=0.536$ to $3$ significant figures. Substitution into the geometric moment formulas gives $E(X)=1/p=1.8662248359\ldots=1.87$ and $\operatorname{Var}(X)=(1-p)/p^2=1.6165703024\ldots=1.62$ , both to $3$ significant figures.
3	$E(X)=1/p$ and $\sqrt{\operatorname{Var}(X)}=\sqrt{1-p}/p$ $\dfrac{1+\sqrt{1-p}}p=6$ Writing $q=1-p$ gives $1/(1-\sqrt q)=6$ $\sqrt q=5/6$ , so $p=11/36$ $E(X)=36/11$ $\operatorname{Var}(X)=900/121$ $P(X\leq3)=1-(1-p)^3=1-(25/36)^3=31031/46656$	7
(7 marks)		7
Notes
Using the geometric moments gives $(1+\sqrt{1-p})/p=6$ . Put $q=1-p$ . Since $1-q=(1-\sqrt q)(1+\sqrt q)$ , the left-hand side reduces to $1/(1-\sqrt q)$ . Hence $\sqrt q=5/6$ , so $q=25/36$ and $p=11/36$ . The moments are $E(X)=36/11$ and $\operatorname{Var}(X)=q/p^2=900/121$ . Finally, $P(X\leq3)=1-P(X>3)=1-q^3=31031/46656$ .
4	$\dfrac{\operatorname{sd}(X)}{E(X)}=\sqrt{1-p}=3/4$ $p=7/16$ $E(X)=16/7$ $\operatorname{Var}(X)=144/49$ $P(X\leq k)\geq0.95$ is equivalent to $(9/16)^k\leq0.05$ For $k=5$ , $(9/16)^5=0.05631\ldots>0.05$ For $k=6$ , $(9/16)^6=0.03168\ldots\leq0.05$ The least integer is $k=6$	8
(8 marks)		8
Notes
For a geometric variable, $\operatorname{sd}(X)/E(X)=[\sqrt{1-p}/p]/(1/p)=\sqrt{1-p}$ . Thus $\sqrt{1-p}=3/4$ , so $p=7/16$ , $E(X)=16/7$ and $\operatorname{Var}(X)=(9/16)/(49/256)=144/49$ . Also $P(X\leq k)=1-(1-p)^k=1-(9/16)^k$ . Since $(9/16)^5=0.0563135\ldots$ and $(9/16)^6=0.0316764\ldots$ , the least suitable integer is $k=6$ .
5	$p_X=1/2$ $\dfrac{1-p_Y}{p_Y^2}=30$ , so $30p_Y^2+p_Y-1=0$ $p_Y=1/6$ $P(X<Y)=\displaystyle\sum_{x=1}^{\infty}\frac12\left(\frac12\right)^{x-1}\left(\frac56\right)^x$ $P(X<Y)=5/7$ $P(X=Y)=\displaystyle\sum_{x=1}^{\infty}\frac1{12}\left(\frac5{12}\right)^{x-1}$ $P(X=Y)=1/7$ $P(X>Y)=1-5/7-1/7=1/7$	8
(8 marks)		8
Notes
$E(X)=1/p_X=2$ gives $p_X=1/2$ . For $Y$ , $(1-p_Y)/p_Y^2=30$ gives $30p_Y^2+p_Y-1=0$ , whose valid root is $p_Y=1/6$ . Independence gives $P(X=x,Y>x)=(1/2)(1/2)^{x-1}(5/6)^x$ , so the resulting geometric series has first term $5/12$ , ratio $5/12$ and sum $5/7$ . Similarly, $P(X=x,Y=x)=(1/12)(5/12)^{x-1}$ , whose sum is $1/7$ . The remaining probability is $1-5/7-1/7=1/7$ , and the three probabilities sum to $1$ .

FS1-3.3 · Mean and variance of negative binomial distribution.

Tier 1 · Easy

Mark scheme for FS1-3.3 Tier 1 · Easy
Question	Scheme	Marks
1	$E(X)=16$ $\operatorname{Var}(X)=48$	2
(2 marks)		2
Notes
Here $r=4$ and $p=0.25$ . Thus $E(X)=r/p=4/0.25=16$ and $\operatorname{Var}(X)=r(1-p)/p^2=4(0.75)/0.25^2=48$ .
2	$\operatorname{Var}(X)=r(1-p)/p^2=20r$ , so $20r=100$ $r=5$ $E(X)=r/p=25$ $P(X=7)=\binom64(0.2)^5(0.8)^2=0.003072$	4
(4 marks)		4
Notes
$\operatorname{Var}(X)=r(1-p)/p^2=r(0.8)/(0.2)^2=20r$ , so $20r=100$ and hence $r=5$ , a positive integer as required. Then $E(X)=r/p=5/0.2=25$ . For $X=7$ , four successes occur in the first six trials and trial $7$ succeeds, giving $\binom64(0.2)^4(0.8)^2(0.2)=15(0.2)^5(0.8)^2=0.003072$ .

Tier 2 · Standard

Mark scheme for FS1-3.3 Tier 2 · Standard
Question	Scheme	Marks
1	$p=\frac13$ $r=5$ $P(X=7)=0.02743$ to $4$ significant figures	6
(6 marks)		6
Notes
The ratio variance/mean is $2=(1-p)/p$ , so $p=1/3$ . Since $r/p=15$ , $r=5$ . Therefore $P(X=7)=\binom64(1/3)^5(2/3)^2=0.0274348\ldots$ .
2	$r$ must be a positive integer and $0<p\leq1$ , so $r\geq2$ $r^2(r-2)=32$ $r=4$ is the only possible value $p=1/2$ $E(X)=r/p=8$ $\operatorname{Var}(X)=r(1-p)/p^2=8$ $P(X=6)=\binom53(1/2)^4(1/2)^2=5/32$	7
(7 marks)		7
Notes
The restrictions on the parameters give integer $r\geq2$ . Substituting $p=2/r$ into $\operatorname{Var}(X)=r(1-p)/p^2=8$ gives $r^2(r-2)=32$ . The left-hand side is strictly increasing for $r\geq2$ , and $r=4$ gives $32$ , so this is the unique possible integer and $p=1/2$ . Therefore $E(X)=r/p=8$ and $\operatorname{Var}(X)=8$ . For $X=6$ , three of the first five trials succeed and trial $6$ succeeds, giving $\binom53(1/2)^6=5/32$ .
3	$P(X\leq2)=0$ and $P(X=3)>0$ , so $r=3$ $\dfrac{\operatorname{Var}(X)}{[E(X)]^2}=\dfrac{1-p}{r}=1/9$ $1-p=r/9=1/3$ , so $p=2/3$ $E(X)=r/p=9/2$ $P(X=3)=p^3=8/27$ $P(X=4)=\binom32p^3(1-p)=3(8/27)(1/3)=8/27$	6
(6 marks)		6
Notes
A negative binomial trial number cannot be below $r$ . The two support statements therefore give $r=3$ . Also $\operatorname{Var}(X)/[E(X)]^2=(1-p)/r$ . Squaring the given standard-deviation relation gives $(1-p)/r=1/9$ , so $1-p=1/3$ and $p=2/3$ . Hence $E(X)=r/p=9/2$ . The third success occurs on trial $3$ with probability $p^3=8/27$ . It occurs on trial $4$ when exactly two of the first three trials succeed and trial $4$ succeeds, so $P(X=4)=\binom32p^3(1-p)=3(8/27)(1/3)=8/27$ as required.

Tier 3 · Hard

Mark scheme for FS1-3.3 Tier 3 · Hard
Question	Scheme	Marks
1	$P(X\leq8)=0.04981$ to $4$ significant figures $P(X=8\mid X\leq8)=\dfrac{189}{304}=0.6217$ to $4$ significant figures $E(X\mid X\leq8)=\dfrac{573}{76}=7.539$ to $4$ significant figures	7
(7 marks)		7
Notes
Sum the negative binomial probabilities for $x=6,7,8$ : $P(X\leq8)=\sum_{x=6}^{8}\binom{x-1}{5}(0.4)^6(0.6)^{x-6}=19456/390625=0.04980736\ldots$ . Also $P(X=8)=\binom75(0.4)^6(0.6)^2=12096/390625$ , so $P(X=8\mid X\leq8)=(12096/390625)/(19456/390625)=189/304$ . Finally $E(X\mid X\leq8)=[\sum_{x=6}^{8}xP(X=x)]/P(X\leq8)=573/76=7.539473\ldots$ .
2	$(1-p)/p=3$ $p=1/4$ $p^r=1/1024=(1/4)^5$ $r=5$ $r=5$ is a positive integer $E(X)=20$ $\operatorname{Var}(X)=60$ $P(X\leq7)=211/16384$	8
(8 marks)		8
Notes
Dividing variance by mean gives $(1-p)/p=3$ , hence $p=1/4$ . The event $X=r$ consists of $r$ consecutive successes, so $p^r=1/1024=(1/4)^5$ and $r=5$ , which is a positive integer. Therefore $E(X)=r/p=20$ and $\operatorname{Var}(X)=r(1-p)/p^2=60$ . Finally $P(X\leq7)=\sum_{x=5}^{7}\binom{x-1}{4}(1/4)^5(3/4)^{x-5}=211/16384$ .
3	$E(F)=\dfrac{r(1-p)}p=7$ $\operatorname{Var}(F)=\dfrac{r(1-p)}{p^2}=14$ $1/p=14/7=2$ , so $p=1/2$ $r=7$ $\dfrac{\operatorname{Var}(F)}{E(F)}=\dfrac1p$ For $0<p<1$ , $1/p>1$ , so $\operatorname{Var}(F)>E(F)$ and equality is impossible $P(F\leq3)=\displaystyle\sum_{f=0}^{3}\binom{f+6}{6}(1/2)^7(1/2)^f$ $P(F\leq3)=11/64$	8
(8 marks)		8
Notes
Since $F=X-r$ , $E(F)=r(1-p)/p$ and $\operatorname{Var}(F)=r(1-p)/p^2$ . Dividing the second given moment by the first gives $1/p=2$ , so $p=1/2$ . Then $r(1-p)/p=r=7$ . In general, for $r>0$ and $0<p<1$ , $E(F)>0$ and $\operatorname{Var}(F)/E(F)=1/p>1$ , so equality is impossible. With the derived parameters, $P(F\leq3)=\sum_{f=0}^{3}\binom{f+6}{6}(1/2)^7(1/2)^f=(8+28+56+84)/1024=11/64$ .