Question 13 - A-Level Maths

AQA Paper 3 2019 June — Question 13 10 marks

Exam Board	AQA
Module	Paper 3 (Paper 3)
Year	2019
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Binomial Distribution
Type	E(X) and Var(X) with probability calculations
Difficulty	Moderate -0.8 This is a straightforward binomial distribution question requiring only standard formula application: mean (np), variance (np(1-p)), and probability calculations using either tables or calculator functions. All parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure and need for careful calculator work.
Spec	2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities 5.02d Binomial: mean np and variance np(1-p)

Patrick is practising his skateboarding skills. On each day, he has 30 attempts at performing a difficult trick. Every time he attempts the trick, there is a probability of 0.2 that he will fall off his skateboard. Assume that the number of times he falls off on any given day may be modelled by a binomial distribution.

1. Find the mean number of times he falls off in a day. [1 mark]
2. Find the variance of the number of times he falls off in a day. [1 mark]
1. Find the probability that, on a particular day, he falls off exactly 10 times. [2 marks]
2. Find the probability that, on a particular day, he falls off 5 or more times. [3 marks]
Patrick has 30 attempts to perform the trick on each of 5 consecutive days.
1. Calculate the probability that he will fall off his skateboard at least 5 times on each of the 5 days. [2 marks]
2. Explain why it may be unrealistic to use the same value of 0.2 for the probability of falling off for all 5 days. [1 mark]

Show mark scheme Show mark scheme source

Question 13:

(a)(i) ---

13

Answer	Marks	Guidance
(a)(i)	Obtains correct mean	1.1b

(a)(ii) ---

13

Answer	Marks	Guidance
(a)(ii)	Obtains correct variance	1.1b

(b)(i) ---

13

Answer	Marks
(b)(i)	Uses the Binomial formula with

n = 30, p = 0.2 or

PI by correct answer

Answer	Marks	Guidance
𝑃𝑃(𝑋𝑋 ≤ 10)−𝑃𝑃(𝑋𝑋 ≤ 9)	1.1a	M1

𝑃𝑃(𝑋𝑋 = 10)= � �0.2 0.8

10 = 0.0355

Obtains correct probability

Answer	Marks	Guidance
AWFW [0.035, 0.036]	1.1b	A1

(b)(ii) ---

13

Answer	Marks
(b)(ii)	Calculates either or

using the Binomial

distribution 𝑃𝑃(𝑋𝑋 ≤ 4) = 0.255

Answer	Marks	Guidance
𝑃𝑃 (𝑋𝑋 ≤ 5)= 0.4275	3.1b	M1

𝑃𝑃(𝑋𝑋 ≥ 5)= 1−𝑃𝑃( 𝑋𝑋 ≤ 4)

= 1−0.255

= 0.745

States or

subtracts their stated value of

Answer	Marks	Guidance
𝑃𝑃( 𝑋𝑋fro≥m5 1) = 1− 𝑃𝑃(𝑋𝑋 ≤ 4)	1.1b	M1

𝑃𝑃O(b𝑋𝑋ta≤ins4 )correct probability

Answer	Marks	Guidance
AWFW [0.74, 0.75]	1.1b	A1

(c)(i) ---

13

Answer	Marks	Guidance
(c)(i)	Raises their 0.745 to power of 5	3.1b

0.745 = 0.229

Obtains their correct probability

FT their

AWRT their

Answer	Marks	Guidance
0.745	1.1b	A1F

(c)(ii) ---

13

Answer	Marks
(c)(ii)	Gives a valid0 r.2e2a9son that

probability/likelihood/chances may

change/increase/decrease as a result

of external factor change over 5 day

Answer	Marks	Guidance
period or Patrick improves	3.5b	E1

Patrick improves

Answer	Marks	Guidance
Total	10
Q	Marking Instructions	AO

Question 13:
--- 13
(a)(i) ---
13
(a)(i) | Obtains correct mean | 1.1b | B1 | 6
--- 13
(a)(ii) ---
13
(a)(ii) | Obtains correct variance | 1.1b | B1 | 4.8
--- 13
(b)(i) ---
13
(b)(i) | Uses the Binomial formula with
n = 30, p = 0.2 or
PI by correct answer
𝑃𝑃(𝑋𝑋 ≤ 10)−𝑃𝑃(𝑋𝑋 ≤ 9) | 1.1a | M1 | 30 10 20
𝑃𝑃(𝑋𝑋 = 10)= � �0.2 0.8
10
= 0.0355
Obtains correct probability
AWFW [0.035, 0.036] | 1.1b | A1
--- 13
(b)(ii) ---
13
(b)(ii) | Calculates either or
using the Binomial
distribution 𝑃𝑃(𝑋𝑋 ≤ 4) = 0.255
𝑃𝑃 (𝑋𝑋 ≤ 5)= 0.4275 | 3.1b | M1 | 𝑃𝑃(𝑋𝑋 ≤ 4)= 0.255
𝑃𝑃(𝑋𝑋 ≥ 5)= 1−𝑃𝑃( 𝑋𝑋 ≤ 4)
= 1−0.255
= 0.745
States or
subtracts their stated value of
𝑃𝑃( 𝑋𝑋fro≥m5 1) = 1− 𝑃𝑃(𝑋𝑋 ≤ 4) | 1.1b | M1
𝑃𝑃O(b𝑋𝑋ta≤ins4 )correct probability
AWFW [0.74, 0.75] | 1.1b | A1
--- 13
(c)(i) ---
13
(c)(i) | Raises their 0.745 to power of 5 | 3.1b | M1 | 5
0.745 = 0.229
Obtains their correct probability
FT their
AWRT their
0.745 | 1.1b | A1F
--- 13
(c)(ii) ---
13
(c)(ii) | Gives a valid0 r.2e2a9son that
probability/likelihood/chances may
change/increase/decrease as a result
of external factor change over 5 day
period or Patrick improves | 3.5b | E1 | Probability may change as
Patrick improves
Total | 10
Q | Marking Instructions | AO | Mark | Typical Solution

Show LaTeX source

Patrick is practising his skateboarding skills. On each day, he has 30 attempts at performing a difficult trick.

Every time he attempts the trick, there is a probability of 0.2 that he will fall off his skateboard.

Assume that the number of times he falls off on any given day may be modelled by a binomial distribution.

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the mean number of times he falls off in a day. [1 mark]

\item Find the variance of the number of times he falls off in a day. [1 mark]
\end{enumerate}

\item \begin{enumerate}[label=(\roman*)]
\item Find the probability that, on a particular day, he falls off exactly 10 times. [2 marks]

\item Find the probability that, on a particular day, he falls off 5 or more times. [3 marks]
\end{enumerate}

\item Patrick has 30 attempts to perform the trick on each of 5 consecutive days.

\begin{enumerate}[label=(\roman*)]
\item Calculate the probability that he will fall off his skateboard at least 5 times on each of the 5 days. [2 marks]

\item Explain why it may be unrealistic to use the same value of 0.2 for the probability of falling off for all 5 days. [1 mark]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2019 Q13 [10]}}

This paper (17 questions)

View full paper

Q1 1 Q2 1 Q3 1 Q4 3 Q5 5 Q6 4 Q7 8 Q8 12 Q9 15 Q10 1 Q11 1 Q12 6 Q13 10 Q14 7 Q15 3 Q16 10 Q17 12