Question 2 - A-Level Maths

OCR MEI S1 — Question 2 8 marks

Exam Board	OCR MEI
Module	S1 (Statistics 1)
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Discrete Probability Distributions
Type	Sequential trials until success
Difficulty	Standard +0.3 This is a straightforward probability distribution problem requiring enumeration of outcomes and basic expectation/variance calculations. Part (i) involves systematic case-counting with independent probability 1/2, while part (ii) applies standard formulas. The setup requires careful reading but the mathematics is routine A-level statistics.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables

2 A couple plan to have at least one child of each sex, after which they will have no more children. However, if they have four children of one sex, they will have no more children. You should assume that each child is equally likely to be of either sex, and that the sexes of the children are independent. The random variable \(X\) represents the total number of girls the couple have.

Show that \(\mathrm { P } ( X = 1 ) = \frac { 11 } { 16 }\). The table shows the probability distribution of \(X\).
\(r\) 0 1 2 3 4
\(\mathrm { P } ( X = r )\) \(\frac { 1 } { 16 }\) \(\frac { 11 } { 16 }\) \(\frac { 1 } { 8 }\) \(\frac { 1 } { 16 }\) \(\frac { 1 } { 16 }\)
Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).

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Question 2:

Part (i)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(P(X=1) = P(g,b)+P(b,g)+P(b,b,g)+P(b,b,b,g)\)	M1	For any two correct fractions; must have correct ref to numbers of boys and girls, not just fractions
\(= \frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} = \frac{11}{16}\)	M1	For all four correct fractions; with no extras
OR: \(P(X=1) = 1 - P(X\neq1) = 1-(P(bbbb)+P(ggb)+P(gggb)+P(gggg))\) \(= 1-\left(\frac{1}{16}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}\right) = \frac{11}{16}\)	A1	NB Answer given; Accept 0.6875, not 0.688. Watch for use of \(B(4, 0.5)\) \(P(X\leq2)=0.6875\) which gets M0M0A0

Part (ii)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(E(X) = \left(0\times\frac{1}{16}\right)+\left(1\times\frac{11}{16}\right)+\left(2\times\frac{1}{8}\right)+\left(3\times\frac{1}{16}\right)+\left(4\times\frac{1}{16}\right)\) \(= 1\frac{3}{8} = 1.375\)	M1, A1	M1 for \(\Sigma rp\) (at least 3 terms correct); A1 CAO; Allow 1.38, not 1.4; Allow 22/16
\(E(X^2) = \left(0\times\frac{1}{16}\right)+\left(1\times\frac{11}{16}\right)+\left(4\times\frac{1}{8}\right)+\left(9\times\frac{1}{16}\right)+\left(16\times\frac{1}{16}\right)\) \(= 2\frac{3}{4} = 2.75\)	M1	For \(\Sigma r^2p\) (at least 3 terms correct); use of \(E(X-\mu)^2\) gets M1 for attempt at \((x-\mu)^2\), should see \((-1.375)^2\), \((-0.375)^2\), \((0.625)^2\), \((1.625)^2\), \((2.625)^2\)
\(\text{Var}(X) = 2\frac{3}{4} - \left(1\frac{3}{8}\right)^2 = \frac{55}{64} = 0.859\)	M1dep, A1	M1dep for their \(E(X)^2\); A1 FT their \(E(X)\) provided \(\text{Var}(X)>0\); 0.86, not 0.9; Using 1.38 gets Var of 0.8456 gets A1

## Question 2:

### Part (i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=1) = P(g,b)+P(b,g)+P(b,b,g)+P(b,b,b,g)$ | M1 | For any two correct fractions; must have correct ref to numbers of boys and girls, not just fractions |
| $= \frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} = \frac{11}{16}$ | M1 | For all four correct fractions; with no extras |
| OR: $P(X=1) = 1 - P(X\neq1) = 1-(P(bbbb)+P(ggb)+P(gggb)+P(gggg))$ $= 1-\left(\frac{1}{16}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}\right) = \frac{11}{16}$ | A1 | *NB Answer given*; Accept 0.6875, not 0.688. Watch for use of $B(4, 0.5)$ $P(X\leq2)=0.6875$ which gets M0M0A0 |

### Part (ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \left(0\times\frac{1}{16}\right)+\left(1\times\frac{11}{16}\right)+\left(2\times\frac{1}{8}\right)+\left(3\times\frac{1}{16}\right)+\left(4\times\frac{1}{16}\right)$ $= 1\frac{3}{8} = 1.375$ | M1, A1 | M1 for $\Sigma rp$ (at least 3 terms correct); A1 CAO; Allow 1.38, not 1.4; Allow 22/16 |
| $E(X^2) = \left(0\times\frac{1}{16}\right)+\left(1\times\frac{11}{16}\right)+\left(4\times\frac{1}{8}\right)+\left(9\times\frac{1}{16}\right)+\left(16\times\frac{1}{16}\right)$ $= 2\frac{3}{4} = 2.75$ | M1 | For $\Sigma r^2p$ (at least 3 terms correct); use of $E(X-\mu)^2$ gets M1 for attempt at $(x-\mu)^2$, should see $(-1.375)^2$, $(-0.375)^2$, $(0.625)^2$, $(1.625)^2$, $(2.625)^2$ |
| $\text{Var}(X) = 2\frac{3}{4} - \left(1\frac{3}{8}\right)^2 = \frac{55}{64} = 0.859$ | M1dep, A1 | M1dep for their $E(X)^2$; A1 FT their $E(X)$ provided $\text{Var}(X)>0$; **0.86, not 0.9**; Using 1.38 gets Var of 0.8456 gets A1 |

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2 A couple plan to have at least one child of each sex, after which they will have no more children. However, if they have four children of one sex, they will have no more children. You should assume that each child is equally likely to be of either sex, and that the sexes of the children are independent. The random variable $X$ represents the total number of girls the couple have.\\
(i) Show that $\mathrm { P } ( X = 1 ) = \frac { 11 } { 16 }$.

The table shows the probability distribution of $X$.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 16 }$ & $\frac { 11 } { 16 }$ & $\frac { 1 } { 8 }$ & $\frac { 1 } { 16 }$ & $\frac { 1 } { 16 }$ \\
\hline
\end{tabular}
\end{center}

(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.

\hfill \mbox{\textit{OCR MEI S1  Q2 [8]}}

This paper (6 questions)

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Q1 7 Q2 8 Q3 7 Q4 7 Q5 7 Q6 8

\(r\)	0	1	2	3	4
\(\mathrm { P } ( X = r )\)	\(\frac { 1 } { 16 }\)	\(\frac { 11 } { 16 }\)	\(\frac { 1 } { 8 }\)	\(\frac { 1 } { 16 }\)	\(\frac { 1 } { 16 }\)