Mathematics 9709/51 — May/June 2020

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

questions

marks

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

Q1ProbabilityDiscrete Random VariablesFree sample

The score when two fair six-sided dice are thrown is the sum of the two numbers on the upper faces.

(a)

Show that the probability that the score is 4 is $\frac{1}{12}$ .

DifficultyEasy

Worked solution

Approach

Count the equally likely ordered outcomes of two dice, then identify which outcomes give a total of 4.

Working

Each die has 6 outcomes, so there are

6 \times 6 = 36

equally likely ordered outcomes.

The outcomes with total 4 are

(1,3), \quad (3,1), \quad (2,2).

Therefore

P(\text{score } 4) = \frac{3}{36} = \frac{1}{12}.

Answer

\frac{1}{12}

Final answer

1/12

Detailed explanation

Walkthrough

We have two distinguishable dice, so there are $6 \times 6 = 36$ equally likely ordered outcomes. To find the probability of scoring 4, list the ordered pairs that add to 4: $(1,3)$ , $(3,1)$ and $(2,2)$ . That is 3 favourable outcomes. Therefore the probability is $\frac{3}{36} = \frac{1}{12}$ .

Key Takeaways

This question tests the idea of a uniform sample space: when all outcomes are equally likely, probability is the number of favourable outcomes divided by the total number of outcomes. It is important to use ordered outcomes because the two dice are physically distinct.

Common Mistakes

Forgetting that $(1,3)$ and $(3,1)$ are two different outcomes.
Counting only the single unordered pair $(1,3)$ and $(2,2)$ , giving only 2 favourable outcomes.
Using a sample space of 21 unordered outcomes without adjusting for unequal probabilities.

Things to Be Careful About

Make sure both dice are treated as distinct. The total number of equally likely outcomes is $6 \times 6 = 36$ , not 21. The outcome $(2,2)$ is counted once, while $(1,3)$ and $(3,1)$ are counted separately.

Techniques used

enumerate ordered outcomes of two dicecount favourable outcomesdivide by total sample space size

(b)

The two dice are thrown repeatedly until a score of 4 is obtained. The number of throws taken is denoted by the random variable $X$ .

Find the mean of $X$ .

DifficultyEasy

Worked solution

Approach

Each throw is an independent trial with the same probability $p$ of obtaining a score of 4. The number of throws until the first success, $X$ , follows a geometric distribution. The mean of this distribution is $\frac{1}{p}$ .

Working

From part (a), the success probability is

p = \frac{1}{12}.

For a geometric distribution,

E(X) = \frac{1}{p}.

E(X) = \frac{1}{\frac{1}{12}} = 12.

Answer

12

Final answer

Detailed explanation

Walkthrough

The random variable $X$ counts the number of throws needed until the first score of 4 appears. Each throw is independent and has the same probability of success $p=\frac{1}{12}$ , so $X$ has a geometric distribution. For a geometric random variable, the expected number of trials until the first success is $\frac{1}{p}$ . Substituting gives $\frac{1}{1/12}=12$ .

Key Takeaways

A geometric distribution models the number of independent trials needed to achieve the first success. Its mean is the reciprocal of the success probability, $E(X)=\frac{1}{p}$ .

Common Mistakes

Trying to use the binomial mean $np$ ; there is no fixed number of trials here.
Forgetting that the successful throw itself is counted in $X$ , so the mean is $1/p$ , not the expected number of failures $q/p$ .

Things to Be Careful About

Use the success probability $p=\frac{1}{12}$ established in part (a). The mean of a geometric distribution is always $\frac{1}{p}$ .

Techniques used

model repeated trials until first success as a geometric distributionapply the mean formula for a geometric distribution

(c)

Find the probability that a score of 4 is first obtained on the 6th throw.

DifficultyEasy

Worked solution

Approach

For $X$ to first equal 4 on the 6th throw, the first five throws must all fail, and the sixth must succeed. Use the geometric distribution formula

P(X = x) = (1-p)^{x-1}p.

Working

Here $p = \frac{1}{12}$ , so $1-p = \frac{11}{12}$ , and $x = 6$ :

P(X=6) = \left(\frac{11}{12}\right)^5 \cdot \frac{1}{12}.

Calculating,

11^5 = 161051, \quad 12^6 = 2985984,

P(X=6) = \frac{161051}{2985984} \approx 0.0539.

Answer

\frac{161051}{2985984} \approx 0.0539

Final answer

161051/2985984 ≈ 0.0539

Detailed explanation

Walkthrough

A score of 4 must first appear on the 6th throw. That means the first five throws are all failures, each with probability $1-p = \frac{11}{12}$ , and the 6th throw is a success with probability $p=\frac{1}{12}$ . Because throws are independent, multiply the probabilities: $\left(\frac{11}{12}\right)^5 \times \frac{1}{12}$ . This is exactly the geometric probability $P(X=6)$ .

Key Takeaways

For a geometric distribution with success probability $p$ , the probability that the first success occurs on the $x$ -th trial is $P(X=x)=(1-p)^{x-1}p$ .

Common Mistakes

Using exponent 6 instead of 5; the success occurs on the 6th throw, so only the preceding 5 throws are failures.
Adding the failure and success probabilities instead of multiplying them.

Things to Be Careful About

The exponent is one less than the trial number because the success is not part of the failures. Keep the final answer as an exact fraction or give a sensible decimal such as 0.0539.

Techniques used

apply the geometric probability mass functioncompute a power and multiply by the success probability

(d)

Find $P(X < 8)$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the complement. The event $X < 8$ means a score of 4 is obtained on one of the first seven throws. The only way $X \ge 8$ is for the first seven throws all to fail. Thus

P(X < 8) = 1 - P(X \ge 8) = 1 - \left(\frac{11}{12}\right)^7.

Working

With $p=\frac{1}{12}$ , the probability of failure on a single throw is

1-p = \frac{11}{12}.

Therefore

P(X < 8) = 1 - \left(\frac{11}{12}\right)^7.

Numerically,

12^7 = 35831808, \quad 11^7 = 19487171,

P(X < 8) = \frac{35831808 - 19487171}{35831808} = \frac{16344637}{35831808} \approx 0.456.

Answer

\frac{16344637}{35831808} \approx 0.456

Final answer

16344637/35831808 ≈ 0.456

Detailed explanation

Walkthrough

We want the probability that the first success occurs before the 8th throw, i.e. $X=1,2,\ldots,7$ . Rather than summing seven geometric probabilities, consider the complement: $X \ge 8$ . This happens exactly when none of the first seven throws is a 4. Each such throw has failure probability $\frac{11}{12}$ , so $P(X \ge 8)=\left(\frac{11}{12}\right)^7$ by independence. Therefore $P(X<8)=1-\left(\frac{11}{12}\right)^7$ , which evaluates to about 0.456.

Key Takeaways

For geometric distributions, probabilities of the form $P(X<k)$ are often easiest to find using the complement, because $P(X\ge k)$ corresponds to $k-1$ consecutive failures. This links the geometric distribution with the concept of independent repeated trials.

Common Mistakes

Writing $P(X < 8) = 1 - \left(\frac{11}{12}\right)^8$ instead of using power 7.
Adding the probabilities $P(X=1)+\cdots+P(X=7)$ without recognising the geometric series shortcut, leading to algebra errors.
Confusing $<8$ with $\le 7$ ; both are the same for integer-valued $X$ , but the exponent in the complement is 7 because $X\ge 8$ means 7 failures before the 8th throw.

Things to Be Careful About

Since $X$ is an integer, $X<8$ is the same as $X\le 7$ . The complement $X\ge 8$ occurs when the first 7 throws all fail, so the power is 7, not 8. Give the exact fraction or a decimal correct to 3 significant figures.

Techniques used

use the complement eventcompute the probability that all first seven trials failevaluate the resulting fraction

The rest of this paper

6 more questions

Q2Permutations and Combinations6M
Q3Probability · Discrete Random Variables7M
Q4Permutations and Combinations4M
Q5Probability8M
Q6The Normal Distribution9M
Q7Representation of Data11M

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