What is S4.1: Discrete distributions and the binomial?

A discrete distribution lists the values with probabilities that are each between 0 and 1 and add to 1; a discrete uniform distribution gives every value the same probability.

How many marks are S4.1 questions worth?

S4.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has S4.1 appeared in?

We have not yet indexed a verified past-paper appearance for S4.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

S4.1

Understand and use simple, discrete probability distributions (mean and variance of discrete random variables excluded), including the binomial distribution as a model; calculate probabilities using the binomial distribution.

Draft — not yet indexed

Discrete distributions and the binomial

Worked answers and methods for S4.1 on Edexcel A-level Maths 9MA0.

Definition

Specification S4.1: understand and use simple, discrete probability distributions (calculation of the mean and variance of discrete random variables is excluded), including the binomial distribution as a model; calculate probabilities using the binomial distribution.

Explanation

A discrete distribution lists the values with probabilities that are each between $0$ and $1$ and add to $1$ ; a discrete uniform distribution gives every value the same probability.
$X\sim\mathrm B(n,p)$ models the number of successes in $n$ trials when the trials are independent, each has two outcomes and the probability of success $p$ is constant; those conditions are the answer to “explain why the binomial is suitable”.
Write the model down, $X\sim\mathrm B(10,\tfrac16)$ , before any calculation.
Point probabilities use $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ and cumulative probabilities come from the calculator, but the inequality must be translated for an integer variable: $P(X<3)=P(X\leq2)$ , $P(X\geq3)=1-P(X\leq2)$ , $P(X>3)=1-P(X\leq3)$ .
A probability found in one part often becomes the $p$ of a second binomial in the next, and $np$ gives an expected count.

Worked example

The random variable

X\sim\mathrm B(14,0.4)

. Find

P(X<5)

, giving your answer to 4 decimal places.

(2)

(Total for Question 1 is 2 marks)

Working

1.For an integer-valued variable, $P(X<5)=P(X\leq4)$ .
2.From the cumulative binomial function with $n=14$ and $p=0.4$ : $P(X\leq4)=0.27925\ldots$ .
3.To 4 decimal places, $P(X<5)=0.2793$ .

Answer: $0.2793$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	$P(X\leq4)$ with $\mathrm B(14,0.4)$ , or a sum of five point probabilities	M1
	awrt $0.2793$	A1
Notes
M1: $P(X\leq5)=0.4859$ scores M0

(2 marks)

Common mistakes

Don't use $P(X<k)$ when the question asks for $P(X\leq k)$ .
Don't calculate a single binomial probability when the event is cumulative, or omit an endpoint from the sum.

Mark-scheme wording to use

“M1 for sight or use of the correct distribution. Must have B, or Bin or Bpd or Bcd and the correct value for n and p, just n = 10, p = 1/6 is M0 … 1st A1 for awrt 0.155” — Pearson 9MA0, June 2024 Paper 31, Q1(a)(i)
“2nd A1 for awrt 0.775” — Pearson 9MA0, June 2024 Paper 31, Q1(a)(ii)
“M1 for selecting an appropriate binomial model T ~ B(10, p), where p is the probability of at least 15 successes from B(24, 0.55). A1 for p = 0.299126..., accepting 0.3 or better. A1 for P(T >= 5) awrt 0.149” — Pearson 9MA0, Sample Paper 31, Q5(b)

What the examiner wants

The inequality read correctly. The June 2024 Paper 31 report says almost all wrote down the right binomial model and found $P(X=3)$ , but for $P(X<3)$ an answer of $0.930$ from $P(X\leq3)$ was often seen, suggesting unfamiliarity with the calculator, and others misread the inequality and calculated $1-P(X\leq2)$ .

Exam tip

Translate the inequality into the exact binomial range before using cumulative probabilities and round only at the end.

Worked practice

Tier 1 · Easy

A discrete random variable

X

takes values

0

1

and

2

with probabilities

k

3k

and

4k

respectively. Find

k

and

P(X\geq1)

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$k=\frac18$ . $P(X\geq1)=\frac78$ .	3
Notes
Probabilities sum to $1$ , so $k+3k+4k=8k=1$ and $k=1/8$ . Therefore $P(X\geq1)=3k+4k=7k=7/8$ .

(3 marks)

Tier 2 · Standard

The random variable

X

has a discrete uniform distribution on

\{-2,0,2,4,6\}

. Write down its probability distribution, then find

P(X<3)

and

P(|X|\geq4)

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$P(X=x)=\dfrac15$ for each $x\in\{-2,0,2,4,6\}$ $P(X<3)=\dfrac35$ $P(\|X\|\geq4)=\dfrac25$	3
Notes
There are five equally likely values, so each has probability $1/5$ . The event $X<3$ contains $-2,0,2$ , giving $3/5$ . The event $\|X\|\geq4$ contains $4$ and $6$ , giving $2/5$ .

(3 marks)

Tier 3 · Hard

A binomial random variable

X

has

10

trials and satisfies

P(X=0)=0.8^{10}

. Find its success probability

p

, then calculate

P(X\geq3)

4

decimal places.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$p=0.2$ . $P(X\geq3)=0.3222$ .	5
Notes
For $X\sim\operatorname{B}(10,p)$ , $P(X=0)=(1-p)^{10}$ . Hence $(1-p)^{10}=0.8^{10}$ , so $p=0.2$ . Then $P(X\geq3)=1-P(X\leq2)=1-[0.8^{10}+10(0.2)(0.8)^9+\binom{10}{2}(0.2)^2(0.8)^8]=0.322200\ldots$ , giving $0.3222$ .

(5 marks)

Tier 1 · Easy

A discrete random variable

Y

takes values

-3

-1

2

and

5

with probabilities

0.10

0.25

0.40

and

0.25

respectively. Find

P(Y>0)

and

P(|Y|\leq2)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$P(Y>0)=0.65$ $P(\|Y\|\leq2)=0.65$	2
Notes
The positive values are $2$ and $5$ , giving $0.40+0.25=0.65$ . The values satisfying $\|Y\|\leq2$ are $-1$ and $2$ , giving $0.25+0.40=0.65$ .

(2 marks)

Tier 2 · Standard

Independent trials each have success probability

0.20

. Find the smallest number of trials needed for the probability of at least one success to exceed

0.95

. State this probability for your minimum number of trials to

4

decimal places.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$14$ trials $P(X\geq1)=0.9560$ when $n=14$	4
Notes
For $n$ trials, $P(X\geq1)=1-0.8^n$ . The requirement is $0.8^n<0.05$ , so $n>\log(0.05)/\log(0.8)=13.425\ldots$ . The smallest integer is $14$ , and $1-0.8^{14}=0.956019\ldots=0.9560$ .

(4 marks)

Tier 3 · Hard

Let

X\sim\operatorname{B}(12,0.25)

. Given that at least one success occurs, find the probability that at least three successes occur. Give your answer to

4

decimal places, using unrounded values in your working.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$0.6293$	5
Notes
Since $X\geq3$ implies $X\geq1$ , the required conditional probability is $P(X\geq3)/P(X\geq1)$ . Now $P(X\geq3)=1-P(X\leq2)=0.609324\ldots$ and $P(X\geq1)=1-P(X=0)=1-0.75^{12}=0.968323\ldots$ . Their ratio is $0.629257\ldots$ , giving $0.6293$ .

(5 marks)

Tier 2 · Standard

Let

X\sim\operatorname{B}(20,0.35)

. Given

P(X\leq3)=0.0443756

P(X\leq4)=0.1181966

and

P(X\leq8)=0.7623776

, find

P(4\leq X\leq8)

and

P(4<X\leq8)

. Give each answer to

4

decimal places.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$P(4\leq X\leq8)=0.7180$ $P(4<X\leq8)=0.6442$	4
Notes
For the inclusive lower endpoint, subtract only values up to $3$ : $P(4\leq X\leq8)=P(X\leq8)-P(X\leq3)=0.7623776-0.0443756=0.7180020$ . For the strict lower endpoint, subtract values up to $4$ : $P(4<X\leq8)=0.7623776-0.1181966=0.6441810$ . Rounding gives $0.7180$ and $0.6442$ .

(4 marks)

Tier 3 · Hard

Eight independent trials share a constant success probability

p

, where

0<p<1

, and

X

denotes their total number of successes. Given that

P(X=2)=3P(X=1)

, find

p

. Hence find

P(X\geq2)

4

decimal places.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$p=\dfrac6{13}$ $P(X\geq2)=0.9445$	6
Notes
For $X\sim\operatorname{B}(8,p)$ , $P(X=2)=28p^2(1-p)^6$ and $P(X=1)=8p(1-p)^7$ . Since $0<p<1$ , division gives $\frac{7p}{2(1-p)}=3$ , so $7p=6(1-p)$ and $p=6/13$ . Therefore $P(X\geq2)=1-P(X=0)-P(X=1)=1-(7/13)^8-8(6/13)(7/13)^7=0.944473\ldots=0.9445$ .

(6 marks)

Tier 3 · Hard

A binomial random variable

X\sim\operatorname{B}(n,p)

satisfies

P(X=1)=6P(X=0)

and

P(X=2)=2P(X=1)

, where

n\geq2

and

0<p<1

. Find

n

and

p

. Hence find the exact value of

P(X\geq2)

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$n=3$ and $p=\dfrac23$ $P(X\geq2)=\dfrac{20}{27}$	6
Notes
Write $q=1-p$ . The first ratio gives $np/q=6$ . Also $P(X=2)/P(X=1)=(n-1)p/(2q)=2$ , so $(n-1)p/q=4$ . Subtracting gives $p/q=2$ , hence $p=2/3$ and then $n=3$ . Therefore $P(X\geq2)=1-P(X=0)-P(X=1)=1-(1/3)^3-3(2/3)(1/3)^2=20/27$ .

(6 marks)

Q10

Tier 3 · Hard

10.

Machine A produces

8

independently inspected components and machine B produces

7

. Every component, regardless of machine, has the same probability

0.40

of being defective, and all inspections are independent. Let

T

be the total number of defective components. State the distribution of

T

and calculate

P(T\geq8)

4

decimal places. Explain why the same binomial distribution for

T

would not be valid if machine B instead had defect probability

0.20

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$T\sim\operatorname{B}(15,0.40)$ $P(T\geq8)=0.2131$ If machine B had probability $0.20$ , the $15$ trials would not share one constant success probability, so their total would not have a binomial distribution with one parameter $p$ .	6
Notes
The two independent counts have the same success probability, so pooling their $8+7=15$ trials gives $T\sim\operatorname{B}(15,0.40)$ . Thus $P(T\geq8)=\sum_{r=8}^{15}\binom{15}{r}(0.4)^r(0.6)^{15-r}=0.213103\ldots=0.2131$ . Different machine probabilities would violate the constant- $p$ condition even if all component outcomes remained independent.

(6 marks)

Verified exam appearances

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Other points in S4 Statistical distributions

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