Two fair coins are thrown at the same time repeatedly until a pair of heads is obtained. The number of throws taken is denoted by the random variable $X$. (a) State the value of $\mathrm{E}(X)$. (b) Find the probability that exactly 5 throws are required to obtain a pair of heads. (c) Find the probability that fewer than 7 throws are required to obtain a pair of heads.

(a) E(X) = 4 · (b) 81/1024 ≈ 0.0791 · (c) 3367/4096 ≈ 0.822

Mathematics 9709/53 — May/June 2023

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

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marks

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

Q1Discrete Random VariablesFree sample

Two fair coins are thrown at the same time repeatedly until a pair of heads is obtained. The number of throws taken is denoted by the random variable $X$ .

(a)

State the value of $\mathrm{E}(X)$ .

DifficultyEasy

Worked solution

Approach

Recognise that $X$ follows a geometric distribution with success probability $p = \mathrm{P}(\text{HH}) = \frac{1}{4}$ . The expectation of a geometric distribution is $\mathrm{E}(X) = \frac{1}{p}$ .

Working

When two fair coins are thrown, the probability of obtaining a pair of heads is:

p = \mathrm{P}(\text{HH}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

Since $X$ is the number of throws until the first success, $X \sim \text{Geo}\left(\frac{1}{4}\right)$ . Therefore:

\mathrm{E}(X) = \frac{1}{p} = \frac{1}{\frac{1}{4}} = 4

Answer

\mathrm{E}(X) = 4

Final answer

E(X) = 4

Detailed explanation

Walkthrough

We first need to identify the probability of success on a single throw. A "success" is getting a pair of heads (HH). Since each coin has two equally likely outcomes, the probability of HH is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ . The random variable $X$ counts how many throws are needed until the first success, which is exactly a geometric distribution with parameter $p = \frac{1}{4}$ . For a geometric distribution, the expected number of trials until the first success is $\mathrm{E}(X) = \frac{1}{p}$ . Substituting $p = \frac{1}{4}$ gives $\mathrm{E}(X) = 4$ . This makes intuitive sense: on average, one in every four throws gives HH, so we expect to need 4 throws.

Key Takeaways

Recognising a geometric distribution when counting trials until the first success.
The expectation formula $\mathrm{E}(X) = \frac{1}{p}$ for a geometric distribution.

Common Mistakes

Forgetting to compute $p = \mathrm{P}(\text{HH})$ first and using $p = \frac{1}{2}$ (the probability of one head).
Confusing $\mathrm{E}(X) = \frac{1}{p}$ with the binomial expectation $np$ .

Things to Be Careful About

The success event is specifically HH, not "at least one head".
$\mathrm{E}(X)$ is the mean number of throws, so it can be a non-integer value; here it happens to be 4.

Techniques used

identify geometric distributionapply expectation formula E(X) = 1/p

(b)

Find the probability that exactly 5 throws are required to obtain a pair of heads.

DifficultyEasy

Worked solution

Approach

For a geometric distribution, $\mathrm{P}(X = k) = (1-p)^{k-1}p$ . Here $p = \frac{1}{4}$ , so $\mathrm{P}(X = 5) = \left(\frac{3}{4}\right)^4 \times \frac{1}{4}$ .

Working

\mathrm{P}(X = 5) = \left(1 - \frac{1}{4}\right)^4 \times \frac{1}{4} = \left(\frac{3}{4}\right)^4 \times \frac{1}{4}

= \frac{81}{256} \times \frac{1}{4} = \frac{81}{1024} \approx 0.0791

Answer

\mathrm{P}(X = 5) = \frac{81}{1024} \approx 0.0791

Final answer

81/1024 ≈ 0.0791

Detailed explanation

Walkthrough

For a geometric distribution, the probability that the first success occurs on the $k$ -th trial is $\mathrm{P}(X = k) = (1-p)^{k-1}p$ . Here $p = \frac{1}{4}$ , so the first 4 throws must all fail (each with probability $\frac{3}{4}$ ) and the 5th throw must succeed (probability $\frac{1}{4}$ ). Thus $\mathrm{P}(X = 5) = \left(\frac{3}{4}\right)^4 \times \frac{1}{4} = \frac{81}{256} \times \frac{1}{4} = \frac{81}{1024} \approx 0.0791$ .

Key Takeaways

The geometric probability formula $\mathrm{P}(X = k) = (1-p)^{k-1}p$ .
Interpreting "exactly 5 throws" as 4 failures followed by 1 success.

Common Mistakes

Using $\left(\frac{3}{4}\right)^5$ instead of $\left(\frac{3}{4}\right)^4$ , forgetting that the 5th throw must be the success.
Mixing up the exponent: it should be $k-1$ failures before the success.

Things to Be Careful About

The success probability $p = \frac{1}{4}$ must be used consistently.
The answer can be given as an exact fraction $\frac{81}{1024}$ or a decimal to 3 significant figures (0.0791).

Techniques used

apply geometric probability formulacompute probability of failures then success

(c)

Find the probability that fewer than 7 throws are required to obtain a pair of heads.

DifficultyMedium-Easy

Worked solution

Approach

Use the complement rule: $\mathrm{P}(X < 7) = 1 - \mathrm{P}(X \geq 7)$ . For a geometric distribution, $\mathrm{P}(X \geq 7)$ is the probability that the first 6 throws all fail, which is $\left(\frac{3}{4}\right)^6$ .

Working

\mathrm{P}(X < 7) = 1 - \left(\frac{3}{4}\right)^6

= 1 - \frac{729}{4096} = \frac{3367}{4096} \approx 0.822

Answer

\mathrm{P}(X < 7) = \frac{3367}{4096} \approx 0.822

Final answer

3367/4096 ≈ 0.822

Detailed explanation

Walkthrough

We want $\mathrm{P}(X < 7)$ , i.e. the probability that the first success occurs on throw 1, 2, 3, 4, 5, or 6. The most efficient way is to use the complement: $\mathrm{P}(X < 7) = 1 - \mathrm{P}(X \geq 7)$ . The event $X \geq 7$ means the first 6 throws are all failures, which has probability $\left(\frac{3}{4}\right)^6$ . Therefore $\mathrm{P}(X < 7) = 1 - \left(\frac{3}{4}\right)^6 = 1 - \frac{729}{4096} = \frac{3367}{4096} \approx 0.822$ . Alternatively, we could sum the geometric probabilities for $k = 1$ to 6, but the complement is much quicker.

Key Takeaways

Using the complement rule to compute cumulative geometric probabilities.
Recognising that $\mathrm{P}(X \geq n) = (1-p)^{n-1}$ for a geometric distribution.

Common Mistakes

Using $\left(\frac{3}{4}\right)^7$ instead of $\left(\frac{3}{4}\right)^6$ for the complement, since $X \geq 7$ requires 6 failures.
Forgetting to subtract from 1, giving the probability of the complement instead.

Things to Be Careful About

"Fewer than 7" means $X < 7$ , i.e. $X = 1, 2, 3, 4, 5, 6$ — not $X \leq 7$ .
The exact answer is $\frac{3367}{4096}$ ; the decimal 0.822 is accepted to at least 3 significant figures.

Techniques used

use complement rulecompute cumulative geometric probability

The rest of this paper

6 more questions

Q2The Normal Distribution5M
Q3Discrete Random Variables7M
Q4Representation of Data8M
Q5Probability7M
Q6The Normal Distribution · Discrete Random Variables10M
Q7Permutations and Combinations9M

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