Probability, binomial distribution and binomial hypothesis testing – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

The random variable $X\sim \mathrm{B}(20,\,0.3)$. Find

(a) $\mathrm{P}(X=5)$ [1]

(b) $\mathrm{P}(X\le 4)$ [1]

(a) awrt 0.179B1
(b) awrt 0.238B1
(c) $\mathrm{P}(X\le 8)-\mathrm{P}(X\le 3)$ (or $0.8867-0.1071$)M1
(c) awrt 0.780A1
Answer: (a) 0.179; (b) 0.238; (c) 0.780
Examiner tip: '$3not included, so subtract $\mathrm{P}(X\le3)$, not $\mathrm{P}(X\le2)$.

2CoreCalculator4 marks

The Venn diagram shows the probabilities associated with three events, $A$, $B$ and $C$.

(a) Show that the events $A$ and $B$ are independent. [2]

(b) Find $\mathrm{P}\big((B\cup C)'\big)$. [1]

(a) finds $\mathrm{P}(A)=0.26+0.14=0.4$ and $\mathrm{P}(B)=0.14+0.16+0.05=0.35$ and attempts $\mathrm{P}(A)\times\mathrm{P}(B)$ (or attempts $\mathrm{P}(A\mid B)=\frac{0.14}{0.35}$ to compare with $\mathrm{P}(A)$, or $\mathrm{P}(B\mid A)=\frac{0.14}{0.4}$ to compare with $\mathrm{P}(B)$)M1
(a) $0.4\times0.35=0.14=\mathrm{P}(A\cap B)$ (or $\mathrm{P}(A\mid B)=0.4=\mathrm{P}(A)$, or $\mathrm{P}(B\mid A)=0.35=\mathrm{P}(B)$), so $A$ and $B$ are independent (conclusion required)A1
(b) $0.26+0.24=0.5$ (accept $1-0.50$ or equivalent)B1
(c) $A$ and $C$, with a reason, e.g. $\mathrm{P}(A\cap C)=0$ / the circles do not overlap / they have no outcomes in commonB1
Answer: (a) $0.4\times0.35=0.14=\mathrm{P}(A\cap B)$; (b) 0.5; (c) $A$ and $C$
Examiner tip: To prove independence you must show $\mathrm{P}(A)\times\mathrm{P}(B)=\mathrm{P}(A\cap B)$ with the numbers and write a conclusion.

3StandardCalculator4 marks

A gardener plants seeds in trays of 12. Each seed either germinates or does not. Past records show that, overall, 85% of these seeds germinate.

The number of seeds in a tray that germinate is to be modelled by a binomial distribution.

(a) State, in context, two assumptions needed for this model to be valid. [2]

(b) Using this model, find the probability that at least 10 seeds in a tray germinate. [2]

(a) the seeds germinate independently of each other / whether one seed germinates does not affect whether another does (must be in context, e.g. mention seeds or germinating)B1
(a) the probability of a seed germinating is the same (0.85) for every seed / for every tray (in context). Do not credit 'fixed number of trials' or 'two outcomes': both are given in the question. Two correct assumptions with no context at all: award B1B0.B1
(b) $X\sim\mathrm{B}(12,0.85)$ and $1-\mathrm{P}(X\le 9)$ or $\mathrm{P}(X=10)+\mathrm{P}(X=11)+\mathrm{P}(X=12)$ (or $\mathrm{P}(Y\le2)$ where $Y\sim\mathrm{B}(12,0.15)$ counts seeds that fail). May be implied by awrt 0.736 from a calculator binomial function.M1
(b) awrt 0.736A1
Answer: (a) seeds germinate independently; probability of germinating is the same (0.85) for each seed; (b) 0.736
Examiner tip: 'At least 10' is $1-\mathrm{P}(X\le 9)$; students often subtract $\mathrm{P}(X\le10)$ and lose both marks.

4StandardCalculator5 marks

A supermarket claims that 25% of its customers use the self-checkout tills. The manager of one branch believes that the proportion at her branch is lower than this. She takes a random sample of 30 customers and finds that 3 of them used the self-checkout tills.

Test, at the 5% level of significance, whether there is evidence to support the manager's belief. State your hypotheses clearly. [5]

$\mathrm{H}_0: p=0.25,\quad \mathrm{H}_1: p<0.25$B1
$X\sim\mathrm{B}(30,\,0.25)$ usedM1
$\mathrm{P}(X\le 3)=$ awrt 0.0374 (or critical region $X\le 3$ with $\mathrm{P}(X\le3)=0.0374$, $\mathrm{P}(X\le4)=0.0979$)A1
compares with 0.05 (or 3 in CR) and draws a correct non-contextual inference: reject $\mathrm{H}_0$ / significantM1
conclusion in context: there is evidence that the proportion of customers using self-checkout at this branch is less than 25% (the manager's belief is supported)A1
Answer: $\mathrm{P}(X\le3)=0.0374<0.05$: reject $\mathrm{H}_0$ – evidence the proportion is lower than 25%
Examiner tip: Hypotheses must use the parameter $p$, and the final A1 needs a non-assertive conclusion in context ('there is evidence that…').

5HardCalculator6 marks

In a large population, 20% of people are left-handed. A researcher wants to test whether the proportion of left-handed people among musicians is different from 20%. She takes a random sample of 40 musicians.

(a) Using a 5% level of significance, find the critical region for a two-tailed test of whether the proportion is different from 20%. The probability in each tail should be as close as possible to, but not more than, 2.5%. [3]

(b) Find the actual significance level of this test. [1]

In her sample, 13 musicians are left-handed.

(a) $X\sim\mathrm{B}(40,0.2)$ with evidence of using cumulative probabilities in a tail, e.g. $\mathrm{P}(X\le2)=0.0079$ or $\mathrm{P}(X\ge14)=0.0194$M1
(a) $X\le 2$A1
(a) $X\ge 14$A1
(b) $0.0079+0.0194=$ awrt 0.0273 (2.73%)B1
(c) 13 is not in the critical region / do not reject $\mathrm{H}_0$M1
(c) insufficient evidence that the proportion of left-handed musicians is different from 20%A1
Answer: (a) $X\le2$ or $X\ge14$; (b) 0.0273; (c) not significant – no evidence the proportion differs from 20%
Examiner tip: For the upper tail use $\mathrm{P}(X\ge14)=1-\mathrm{P}(X\le13)$ – an off-by-one error here loses the A1.

6HardCalculator6 marks

On any working day the probability that Dev's bus is late is 0.4, independently of other days. A fortnight contains 10 working days.

(a) Find the probability that the bus is late on at least 5 days in a fortnight. [2]

(b) Find the probability that, in the next 6 fortnights, there are exactly 2 fortnights in which the bus is late on at least 5 days. [3]

(a) $X\sim\mathrm{B}(10,0.4)$ and $1-\mathrm{P}(X\le 4)$ (or $\mathrm{P}(X\ge5)$ found directly on calculator)M1
(a) awrt 0.367A1
(b) $Y\sim\mathrm{B}(6,\,p)$ with $p$ = their (a)M1
(b) $\binom{6}{2}p^2(1-p)^4$ or $15p^2(1-p)^4$ with their $p$, or $\mathrm{P}(Y=2)$ from a calculator binomial pdf with $n=6$ and their $p$; may be implied by awrt 0.324M1
(b) awrt 0.324A1
(c) a reason in context why lateness on one day makes lateness on another more (or less) likely, e.g. roadworks, bad weather, a breakdown or a traffic problem can last several days, so if the bus is late one day it is more likely to be late the next. Accept 'same driver/bus each day' only if linked to lateness carrying over (e.g. a slow driver or faulty bus is late every day). Reject just 'days are not independent' or 'same driver' with no link to lateness.B1
Answer: (a) 0.367; (b) 0.324; (c) e.g. roadworks/bad weather can last several days, so lateness on consecutive days is linked
Examiner tip: Spot the two-stage model: part (b) needs a new binomial whose 'success' probability is your answer to (a).

7StandardCalculator6 marks

The discrete random variable $X$ has probability distribution $$\mathrm{P}(X=x)=kx^2,\qquad x=1,2,3$$

(a) Show that $k=\frac{1}{14}$. [1]

(b) Find $\mathrm{P}(X\ge 2)$. [1]

A game consists of observing one value of $X$. The game is played 5 times and the values are independent.

(a) $k(1+4+9)=1\Rightarrow 14k=1$, so $k=\frac{1}{14}$ (shown)B1
(b) $\frac{4}{14}+\frac{9}{14}=\frac{13}{14}$ (awrt 0.929)B1
(c) $\mathrm{P}(X=3)=\frac{9}{14}$B1
(c) $Y\sim\mathrm{B}\!\left(5,\ \text{their }\tfrac{9}{14}\right)$ used, where $Y$ is the number of games with $X=3$M1
(c) $1-\mathrm{P}(Y=0)-\mathrm{P}(Y=1)=1-\left(\frac{5}{14}\right)^5-5\left(\frac{9}{14}\right)\left(\frac{5}{14}\right)^4$ (correct structure with their $p$; allow $1-\mathrm{P}(Y\le1)$ from calculator, or $\mathrm{P}(Y=2)+\dots+\mathrm{P}(Y=5)$)M1
(c) awrt 0.942A1
Answer: (a) shown: $14k=1$; (b) $\frac{13}{14}$; (c) 0.942
Examiner tip: 'At least 2' means subtract both $\mathrm{P}(Y=0)$ and $\mathrm{P}(Y=1)$ from 1.

8StretchCalculator8 marks

A manufacturer of light bulbs states that 10% of its bulbs are faulty.

(a) A quality inspector selects a random sample of $n$ bulbs. Find the smallest value of $n$ for which the probability that none of the bulbs is faulty is less than 0.01. [3]

After a change in the production process, the manufacturer claims the proportion of faulty bulbs has decreased. A random sample of 50 bulbs is taken and 1 is found to be faulty.

(b) Test, at the 5% level of significance, whether there is evidence to support the manufacturer's claim. State your hypotheses clearly. [5]

(a) $0.9^n<0.01$M1
(a) $n>\frac{\ln 0.01}{\ln 0.9}$ $(=43.7\ldots)$, or trial of $0.9^{43}=0.0108$ and $0.9^{44}=0.0097$M1
(a) $n=44$A1
(b) $\mathrm{H}_0:p=0.1,\quad \mathrm{H}_1:p<0.1$B1
(b) $X\sim\mathrm{B}(50,0.1)$ and attempts $\mathrm{P}(X\le1)$M1
(b) awrt 0.0338A1
(b) $0.0338<0.05$ so reject $\mathrm{H}_0$ (significant)M1
(b) there is evidence that the proportion of faulty bulbs has decreased (supports the claim)A1
Answer: (a) $n=44$; (b) $\mathrm{P}(X\le1)=0.0338<0.05$ – evidence the proportion has decreased
Examiner tip: Dividing by $\ln 0.9$ (a negative number) flips the inequality – check your $n$ by substituting back.

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