Mathematics 9709/51 — October/November 2021

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

questions

marks

minutes

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

Q1Discrete Random VariablesFree sample

Two fair coins are thrown at the same time. The random variable $X$ is the number of throws of the two coins required to obtain two tails at the same time.

(a)

Find the probability that two tails are obtained for the first time on the 7th throw.

DifficultyMedium-Easy

Worked solution

Approach

Each throw of the two coins has probability $p = \frac{1}{4}$ of showing two tails, and probability $1-p = \frac{3}{4}$ of not showing two tails. For the first two tails to occur on the 7th throw, the first 6 throws must all fail and the 7th must succeed.

Working

The required probability is:

(1-p)^6 p = \left(\frac{3}{4}\right)^6 \left(\frac{1}{4}\right)

Evaluate:

\left(\frac{3}{4}\right)^6 = \frac{729}{4096}

So:

\frac{729}{4096} \times \frac{1}{4} = \frac{729}{16384}

Answer

P(X = 7) = \frac{729}{16384} \approx 0.0445

Final answer

729/16384 ≈ 0.0445

Detailed explanation

Walkthrough

Each throw of two coins is an independent trial. There are four equally likely outcomes: HH, HT, TH and TT, so the probability of two tails on any one throw is $\frac{1}{4}$ . The probability of not getting two tails is therefore $\frac{3}{4}$ .

The event "two tails are obtained for the first time on the 7th throw" means the first six throws all fail and the seventh throw succeeds. Because the throws are independent, multiply the probabilities:

\left(\frac{3}{4}\right)^6 \times \frac{1}{4}

This is exactly the geometric distribution formula $P(X = n) = (1-p)^{n-1}p$ with $p = \frac{1}{4}$ and $n = 7$ .

Key Takeaways

Recognise repeated independent trials continued until the first success as a geometric distribution. The probability that the first success occurs on the $n$ th trial is $q^{n-1}p$ , where $q = 1-p$ is the probability of failure on one trial.

Common Mistakes

Forgetting the six failures before the success and writing $\frac{1}{4}$ alone.
Using $\left(\frac{3}{4}\right)^7 \times \frac{1}{4}$ instead of exponent 6.
Confusing this with a binomial probability; binomial counts successes in a fixed number of trials, whereas this question asks for the trial number of the first success.

Things to Be Careful About

"At the same time" means each throw of the two coins is one trial; only the outcome TT counts as success. Keep the exact fraction $\frac{729}{16384}$ or give the decimal $0.0445$ to 3 significant figures. The mark scheme requires the form $(1-p)^6 p$ for the method mark.

Techniques used

model the number of throws until the first success with a geometric distributionmultiply the probabilities of independent trials

(b)

Find the probability that it takes more than 9 throws to obtain two tails for the first time.

DifficultyMedium-Easy

Worked solution

Approach

"More than 9 throws" means that no two tails are obtained in the first 9 throws. Each throw independently fails with probability $\frac{3}{4}$ , so the required probability is $\left(\frac{3}{4}\right)^9$ . This is also the sum of the geometric probabilities for first success on throw 10, 11, 12, ...

Working

P(X > 9) = \left(\frac{3}{4}\right)^9

Evaluate:

\left(\frac{3}{4}\right)^9 = \frac{19683}{262144}

Answer

P(X > 9) = \frac{19683}{262144} \approx 0.0751

Final answer

19683/262144 ≈ 0.0751

Detailed explanation

Walkthrough

We need the probability that it takes more than 9 throws, i.e. $P(X > 9)$ . This happens exactly when the first 9 throws all fail to produce two tails. Each throw has failure probability $\frac{3}{4}$ , and the throws are independent, so multiply these probabilities:

\left(\frac{3}{4}\right)^9

Equivalently, sum the geometric probabilities for first success on throw 10, 11, 12, ...:

\sum_{n=10}^{\infty} \left(\frac{3}{4}\right)^{n-1} \frac{1}{4} = \left(\frac{3}{4}\right)^9

The infinite sum is easier to evaluate directly as the tail probability $q^9$ .

Key Takeaways

For a geometric distribution, $P(X > n) = q^n$ , where $q$ is the probability of failure on one trial. Tail probabilities can often be found directly by requiring all of the first $n$ trials to fail.

Common Mistakes

Using $P(X = 9) = \left(\frac{3}{4}\right)^8 \times \frac{1}{4}$ instead of the tail probability.
Using exponent 8 for "more than 9 throws"; the correct exponent is 9 because the first 9 throws must all fail.
Trying to sum infinitely many terms without recognising the geometric tail simplifies to $q^9$ .

Things to Be Careful About

"More than 9 throws" means 10 or more throws are needed, so the first 9 throws must all be failures. The exact fraction is $\frac{19683}{262144}$ , and the decimal is $0.0751$ to 3 significant figures. The mark scheme accepts the form $\left(\frac{3}{4}\right)^n$ with appropriate $n$ for the method mark.

Techniques used

use the tail probability of a geometric distributioncompute the probability that all initial trials fail

The rest of this paper

6 more questions

Q2Representation of Data4M
Q3Probability5M
Q4Discrete Random Variables6M
Q5Permutations and Combinations · Probability10M
Q6Representation of Data10M
Q7The Normal Distribution · Discrete Random Variables11M

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