Question 7 - A-Level Maths

Edexcel S1 2022 October — Question 7 14 marks

Exam Board	Edexcel
Module	S1 (Statistics 1)
Year	2022
Session	October
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Discrete Probability Distributions
Type	Conditional probability with random variables
Difficulty	Standard +0.3 This is a straightforward S1 question testing basic probability distribution concepts: conditional probability (part a), expectation and variance calculations (parts b-d), and independent events (parts e-f). All parts follow standard algorithms with no novel problem-solving required. Part (f) requires systematic enumeration of products but is routine for this level. Slightly above average difficulty only due to the multi-part nature and part (e) requiring careful consideration of all cases.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables 5.02c Linear coding: effects on mean and variance

Adana selects one number at random from the distribution of \(X\) which has the following probability distribution.

\(x\)	0	5	10
\(\mathrm { P } ( X = x )\)	0.1	0.2	0.7

Given that the number selected by Adana is not 5 , write down the probability it is 0
Show that \(\mathrm { E } \left( X ^ { 2 } \right) = 75\)
Find \(\operatorname { Var } ( X )\)
Find \(\operatorname { Var } ( 4 - 3 X )\) Bruno and Charlie each independently select one number at random from the distribution of \(X\)
Find the probability that the number Bruno selects is greater than the number Charlie selects. Devika multiplies Bruno's number by Charlie's number to obtain a product, \(D\)
Determine the probability distribution of \(D\)

Show mark scheme Show mark scheme source

Question 7:

Part (a):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(\left[\frac{0.1}{0.8}\right] = \mathbf{\frac{1}{8}}\)	B1	\(\frac{1}{8}\) oe; allow 0.125; do not ISW

Part (b):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\([0^2\times0.1+]\ 5^2\times0.2 + 10^2\times0.7 = 75^*\)	B1*cso	Correct expression and 75 with no errors seen. Allow \(5^2\times0.2 + 10^2\times0.7 = 75\)

Part (c):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(E(X) = [0\times0.1+]\ 5\times0.2 + 10\times0.7\ [=8]\)	M1	Correct method to find mean; if no method seen award if 8 is seen
\(\text{Var}(X) = 75 - (\text{'8'})^2\)	M1	Attempt at expression for variance: \(75 - (E(X))^2\)
\(\text{Var}(X) = \mathbf{11}\)	A1	11 cao

Part (d):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(\text{Var}(4-3X) = 3^2 \times \text{Var}(X)[= 3^2 \times \text{"11"}]\)	M1	Use of \((-3)^2 \times \text{Var}(X)\) (condone 3 rather than \(-3\) and missing bracket if final answer correct)
\(= \mathbf{99}\)	A1ft	99 or ft \(9\times\)'their (c)'

Part (e):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(P((0,5),(0,10),(5,10))= 0.1\times0.2+0.1\times0.7+0.2\times0.7\ [=0.02+0.07+0.14]= \mathbf{0.23}\)	M1 M1 A1	\(1^{st}\) M1: at least one correct product; \(2^{nd}\) M1: fully correct expression e.g. \(0.1\times0.2+0.3\times0.7\); A1: 0.23 oe
OR: \(P((0,0),(5,5),(10,10))= 1-(0.1^2+0.2^2+0.7^2)= 0.5(1-(0.01+0.04+0.49))= \mathbf{0.23}\)	M1 M1 A1	Alternative method giving same marks

Part (f):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Products: 0, 25, 50, 100	B1	All 4 correct products with no incorrect extras (unless probability 0 associated)
\(P(D=0)=0.1+0.1-0.1\times0.1[=0.19]\); \(P(D=25)=0.2^2\)	M1 M1	\(1^{st}\) M1: correct method for 1 of 4 probabilities; \(2^{nd}\) M1: correct method for 3 of 4 probabilities if total = 1, must be associated with correct product
\(P(D=50)=2\times0.2\times0.7\); \(P(D=100)=0.7^2\)		ALT \(P(D=0)=0.1\times0.1+2\times0.1\times0.2+2\times0.1\times0.7[=0.19]\)
\(D\)	\(0\)	\(25\)
\(P(D=d)\)	0.19	0.04
A1: all four correct probabilities (oe) associated with correct products

# Question 7:

## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left[\frac{0.1}{0.8}\right] = \mathbf{\frac{1}{8}}$ | B1 | $\frac{1}{8}$ oe; allow 0.125; do not ISW |

## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[0^2\times0.1+]\ 5^2\times0.2 + 10^2\times0.7 = 75^*$ | B1*cso | Correct expression and 75 with no errors seen. Allow $5^2\times0.2 + 10^2\times0.7 = 75$ |

## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(X) = [0\times0.1+]\ 5\times0.2 + 10\times0.7\ [=8]$ | M1 | Correct method to find mean; if no method seen award if 8 is seen |
| $\text{Var}(X) = 75 - (\text{'8'})^2$ | M1 | Attempt at expression for variance: $75 - (E(X))^2$ |
| $\text{Var}(X) = \mathbf{11}$ | A1 | 11 cao |

## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Var}(4-3X) = 3^2 \times \text{Var}(X)[= 3^2 \times \text{"11"}]$ | M1 | Use of $(-3)^2 \times \text{Var}(X)$ (condone 3 rather than $-3$ and missing bracket if final answer correct) |
| $= \mathbf{99}$ | A1ft | 99 or ft $9\times$'their (c)' |

## Part (e):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P((0,5),(0,10),(5,10))= 0.1\times0.2+0.1\times0.7+0.2\times0.7\ [=0.02+0.07+0.14]= \mathbf{0.23}$ | M1 M1 A1 | $1^{st}$ M1: at least one correct product; $2^{nd}$ M1: fully correct expression e.g. $0.1\times0.2+0.3\times0.7$; A1: 0.23 oe |
| OR: $P((0,0),(5,5),(10,10))= 1-(0.1^2+0.2^2+0.7^2)= 0.5(1-(0.01+0.04+0.49))= \mathbf{0.23}$ | M1 M1 A1 | Alternative method giving same marks |

## Part (f):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Products: 0, 25, 50, 100 | B1 | All 4 correct products with no incorrect extras (unless probability 0 associated) |
| $P(D=0)=0.1+0.1-0.1\times0.1[=0.19]$; $P(D=25)=0.2^2$ | M1 M1 | $1^{st}$ M1: correct method for 1 of 4 probabilities; $2^{nd}$ M1: correct method for 3 of 4 probabilities if total = 1, must be associated with correct product |
| $P(D=50)=2\times0.2\times0.7$; $P(D=100)=0.7^2$ | | ALT $P(D=0)=0.1\times0.1+2\times0.1\times0.2+2\times0.1\times0.7[=0.19]$ |

| $D$ | $0$ | $25$ | $50$ | $100$ |
|---|---|---|---|---|
| $P(D=d)$ | **0.19** | **0.04** | **0.28** | **0.49** |

A1: all four correct probabilities (oe) associated with correct products |

Show LaTeX source

\begin{enumerate}
  \item Adana selects one number at random from the distribution of $X$ which has the following probability distribution.
\end{enumerate}

\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 0 & 5 & 10 \\
\hline
$\mathrm { P } ( X = x )$ & 0.1 & 0.2 & 0.7 \\
\hline
\end{tabular}
\end{center}

(a) Given that the number selected by Adana is not 5 , write down the probability it is 0\\
(b) Show that $\mathrm { E } \left( X ^ { 2 } \right) = 75$\\
(c) Find $\operatorname { Var } ( X )$\\
(d) Find $\operatorname { Var } ( 4 - 3 X )$

Bruno and Charlie each independently select one number at random from the distribution of $X$\\
(e) Find the probability that the number Bruno selects is greater than the number Charlie selects.

Devika multiplies Bruno's number by Charlie's number to obtain a product, $D$\\
(f) Determine the probability distribution of $D$

\hfill \mbox{\textit{Edexcel S1 2022 Q7 [14]}}

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