Mathematics 9709/52 — February/March 2022

Cambridge A-Level · Probability & Statistics 1 · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Probability · Discrete Random Variables · Representation of Data · The Normal Distribution · Permutations and Combinations

Q1ProbabilityDiscrete Random VariablesFree sample

A fair red spinner has edges numbered 1, 2, 2, 3. A fair blue spinner has edges numbered $-3, -2, -1, -1$ . Each spinner is spun once and the number on the edge on which each spinner lands is noted. The random variable $X$ denotes the sum of the resulting two numbers.

(a)

Draw up the probability distribution table for $X$ .

DifficultyMedium-Easy

Worked solution

Approach

Each spinner has 4 edges and repeated numbers are separate edges, so there are $4 \times 4 = 16$ equally likely pairs. For each possible sum, count the number of pairs giving that sum, then divide by 16.

Working

Counting the 16 pairs:

$X = -2$ : $1 + (-3)$ gives 1 outcome.
$X = -1$ : $1 + (-2)$ and $2 + (-3)$ using the two red 2s give 3 outcomes.
$X = 0$ : $1 + (-1)$ in 2 ways, $2 + (-2)$ in 2 ways, $3 + (-3)$ in 1 way, giving 5 outcomes.
$X = 1$ : $2 + (-1)$ in 4 ways and $3 + (-2)$ in 1 way, giving 5 outcomes.
$X = 2$ : $3 + (-1)$ in 2 ways, giving 2 outcomes.

Hence the probability distribution table is:

$X$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x)$	$\frac{1}{16}$	$\frac{3}{16}$	$\frac{5}{16}$	$\frac{5}{16}$	$\frac{2}{16}$

Check:

\frac{1+3+5+5+2}{16} = \frac{16}{16} = 1

Answer

The probability distribution table for $X$ is:

$X$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x)$	$\frac{1}{16}$	$\frac{3}{16}$	$\frac{5}{16}$	$\frac{5}{16}$	$\frac{2}{16}$

Final answer

X = -2: 1/16, X = -1: 3/16, X = 0: 5/16, X = 1: 5/16, X = 2: 2/16

Detailed explanation

Walkthrough

Since each spinner has four edges, there are $4 \times 4 = 16$ equally likely outcomes when the two spinners are spun. Do not simplify the two red edges labelled 2 or the two blue edges labelled $-1$ away: they are distinct edges, so they each correspond to an equally likely outcome. Work through the sums in order and count outcomes. For example, the sum $1$ arises in $2 \times 2 = 4$ ways because the red 2 can be either of two edges and the blue $-1$ can be either of two edges, plus one more way from $3 + (-2)$ . Dividing the counts by 16 gives the probabilities; checking that they total 1 confirms the table is complete.

Key Takeaways

A probability distribution table lists every possible value of the random variable exactly once, with probabilities between 0 and 1 that sum to 1. When outcomes are equally likely, the probability of a value is frequency divided by the total number of outcomes. Repeated labels still count as separate outcomes.

Common Mistakes

Forgetting that the duplicate 2s and duplicate $-1$ s make some sums more likely than others.
Listing the same $X$ value multiple times or omitting a possible sum.
Using a denominator smaller than 16 or giving probabilities that do not sum to 1.
Treating repeated labels as a single outcome.

Things to Be Careful About

The mark scheme requires correct probabilities linked with correct outcomes. Make it clear where each frequency comes from by counting edge pairs. It also requires no repeated $X$ values in the table. Finally, always check $\sum P(X=x) = 1$ ; this catches any missed outcomes.

Techniques used

enumerate all equally likely outcomesgroup outcomes by their sumdivide frequencies by the total number of outcomes

(b)

Given that $E(X) = 0.25$ , find the value of $\text{Var}(X)$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the variance formula for a discrete random variable:

\text{Var}(X) = E(X^2) - \left(E(X)\right)^2

The value $E(X) = 0.25 = \frac{1}{4}$ is given, so first compute $E(X^2)$ from the probability distribution in part (a).

Working

E(X^2) = \sum x^2 P(X=x)

E(X^2) = \frac{1 \times (-2)^2}{16} + \frac{3 \times (-1)^2}{16} + \frac{5 \times 0^2}{16} + \frac{5 \times 1^2}{16} + \frac{2 \times 2^2}{16}

E(X^2) = \frac{4 + 3 + 0 + 5 + 8}{16} = \frac{20}{16} = \frac{5}{4}

Therefore:

\text{Var}(X) = \frac{5}{4} - \left(\frac{1}{4}\right)^2 = \frac{20}{16} - \frac{1}{16} = \frac{19}{16} = 1.1875

Answer

\text{Var}(X) = \frac{19}{16} = 1.1875

Final answer

Var(X) = 19/16 = 1.1875

Detailed explanation

Walkthrough

Variance measures the spread of $X$ about its mean. The formula is $\text{Var}(X) = E(X^2) - \left(E(X)\right)^2$ . Because $E(X)=0.25$ is given, calculate $E(X^2)$ by weighting each value of $X^2$ by its probability. Important: use $x^2$ , not $x$ , so the negative values $-2$ and $-1$ contribute $4$ and $1$ respectively. Substitute the probabilities from part (a) and simplify. Then subtract $(0.25)^2 = \frac{1}{16}$ . Using fractions makes the final value $\frac{19}{16}$ or $1.1875$ .

Key Takeaways

Variance is not the average of the values; it is the expected squared deviation from the mean, computed as $E(X^2)-[E(X)]^2$ . Squaring each value before weighting is crucial. Exact fractions are often easier than decimals.

Common Mistakes

Calculating $\sum xP(X=x)$ instead of $\sum x^2P(X=x)$ .
Forgetting to subtract the square of the mean, or subtracting $E(X)$ instead of $[E(X)]^2$ .
Using $0.25$ inside the squared term incorrectly, e.g. subtracting $0.25$ instead of $0.0625$ .
Mis-adding fractions by using incompatible denominators.

Things to Be Careful About

Use the table from part (a) correctly, especially the two $5/16$ probabilities. The random variable can be negative, but squaring removes the sign. If a probability table error is followed through correctly, the mark scheme may still award the method mark; write the variance formula explicitly to show it.

Techniques used

use the variance formula for a discrete random variablecompute E(X^2) from the probability distributionsubtract the square of the given mean

The rest of this paper

5 more questions

Q2Discrete Random Variables6M
Q3Representation of Data6M
Q4The Normal Distribution · Probability11M
Q5Permutations and Combinations10M
Q6Discrete Random Variables · Probability12M

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