Discrete distributions and the binomial distribution – 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 $\mathrm{P}(X\le2)$.

2StandardCalculator4 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.

3HardCalculator6 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).

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4StandardCalculator6 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.

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