It is known that, on average, 1 in 300 flowers of a certain kind are white. A random sample of 200 flowers of this kind is selected. (a) Use an appropriate approximating distribution to find the probability that more than 1 flower in the sample is white. (b) Justify the approximating distribution used in part (a). (c) The probability that a randomly chosen flower of another kind is white is 0.02. A random sample of 150 of these flowers is selected. Use an appropriate…

(a) 0.144 · (b) n = 200 > 50 and np = 2/3 < 5 · (c) 0.501

Mathematics 9709/63 — October/November 2020

Cambridge A-Level · Probability & Statistics 2 · worked solutions for every part, with the mark scheme

questions

marks

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Topics The Poisson Distribution · Linear Combinations of Random Variables · Sampling and Estimation · Hypothesis Tests · Continuous Random Variables

Q1The Poisson DistributionLinear Combinations of Random VariablesFree sample

It is known that, on average, 1 in 300 flowers of a certain kind are white. A random sample of 200 flowers of this kind is selected.

(a)

Use an appropriate approximating distribution to find the probability that more than 1 flower in the sample is white.

DifficultyMedium-Easy

Worked solution

Approach

Let $X$ be the number of white flowers in the sample. Exactly, $X \sim \text{B}(200, \frac{1}{300})$ , but because $n$ is large and $p$ is small we approximate $X$ by a Poisson distribution with mean

\lambda = np = 200 \times \frac{1}{300} = \frac{2}{3}

We need $P(X > 1)$ . It is easier to use the complement: subtract the probabilities of $X = 0$ and $X = 1$ from 1.

Working

Using $P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!}$ with $\lambda = \frac{2}{3}$ ,

P(X > 1) = 1 - P(X = 0) - P(X = 1)

= 1 - e^{-\frac{2}{3}} - e^{-\frac{2}{3}}\left(\frac{2}{3}\right)

= 1 - e^{-\frac{2}{3}}\left(1 + \frac{2}{3}\right)

= 0.144 \quad (3\text{ s.f.})

Answer

0.144

Final answer

0.144

Detailed explanation

Walkthrough

Start by recognising the exact model. Each flower is either white or not white, and the sample is random, so the number of white flowers has a binomial distribution $X \sim \text{B}(200, \frac{1}{300})$ . Because $n$ is large and $p$ is very small, the Poisson approximation is appropriate. The mean of the approximating Poisson distribution is $\lambda = np = \frac{2}{3}$ .

The event "more than 1 flower is white" is $X > 1$ . Rather than summing infinitely many probabilities, use the complement: $P(X > 1) = 1 - P(X = 0) - P(X = 1)$ . Substitute the Poisson formula $P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!}$ with $\lambda = \frac{2}{3}$ .

This gives $1 - e^{-\frac{2}{3}}(1 + \frac{2}{3})$ , which evaluates to $0.144$ to 3 significant figures.

Key Takeaways

The Poisson distribution can approximate a binomial distribution when $n$ is large and $p$ is small. The mean is preserved: $\lambda = np$ . For "more than" questions, the complement $1 - P(X \le 1)$ is often quicker than adding infinitely many terms.

Common Mistakes

Forgetting to subtract both $P(X = 0)$ and $P(X = 1)$ ; subtracting only $P(X = 0)$ gives the probability of at least 1, not more than 1.
Using the binomial distribution directly. The mark scheme gives only special-consideration credit (B1) for a correct final answer obtained by binomial or with no working shown.
Rounding $\lambda$ too early; use $\frac{2}{3}$ or at least 3 significant figures.

Things to Be Careful About

The required probability is strictly greater than 1, so $X = 1$ must be included in the complement.
Show the formula and the substitution clearly to earn the method mark.
Give the final answer to 3 significant figures.

Techniques used

approximate a binomial distribution by a Poisson distributionuse the complement rule for a tail probabilitycalculate Poisson probabilities for 0 and 1

(b)

Justify the approximating distribution used in part (a).

DifficultyEasy

Worked solution

Approach

The Poisson approximation to the binomial distribution is appropriate when $n$ is large and $p$ is small. Check the numerical conditions for this sample.

Working

Here $n = 200$ and $p = \frac{1}{300}$ .

n = 200 > 50

and

np = 200 \times \frac{1}{300} = \frac{2}{3} < 5

Equivalently, $p = \frac{1}{300} < 0.1$ . Therefore the binomial distribution is well approximated by $\text{Po}\left(\frac{2}{3}\right)$ .

Answer

$n = 200 > 50 \text{ and } np = \frac{2}{3} < 5$

Final answer

n = 200 > 50 and np = 2/3 < 5

Detailed explanation

Walkthrough

The approximation in part (a) is justified by checking the standard conditions for using a Poisson distribution in place of a binomial distribution. We need both a large sample size and a small probability of success. Here $n = 200 > 50$ and $np = \frac{2}{3} < 5$ ; equivalently $p = \frac{1}{300} < 0.1$ . These numerical checks show the binomial probabilities are very close to Poisson probabilities with mean $\frac{2}{3}$ .

Key Takeaways

A Poisson approximation is not justified by saying " $n$ is large and $p$ is small" without numbers. The standard checks are $n > 50$ and $np < 5$ (or $p < 0.1$ ).

Common Mistakes

Writing only " $n$ is large and $p$ is small". The mark scheme does not accept this vague statement.
Confusing the condition on $p$ with the condition on $np$ ; either can be used, but the values must be stated clearly.

Things to Be Careful About

Use the actual values from the question: $n = 200$ , $p = \frac{1}{300}$ , $np = \frac{2}{3}$ .
The condition can be expressed as $np < 5$ or $p < 0.1$ ; both are acceptable if stated clearly.

Techniques used

state the numerical conditions for using a Poisson approximationcheck n > 50 and np < 5

(c)

The probability that a randomly chosen flower of another kind is white is 0.02. A random sample of 150 of these flowers is selected.

Use an appropriate approximating distribution to find the probability that the total number of white flowers in the two samples is less than 4.

DifficultyMedium

Worked solution

Approach

Let $Y_1$ be the number of white flowers in the first sample and $Y_2$ the number in the second sample. Each is binomial and is approximated by a Poisson distribution:

Y_1 \sim \text{Po}\left(\frac{2}{3}\right), \qquad Y_2 \sim \text{Po}(3)

because $200 \times \frac{1}{300} = \frac{2}{3}$ and $150 \times 0.02 = 3$ . Since the samples are independent, the total $T = Y_1 + Y_2$ is also Poisson with mean $\frac{2}{3} + 3 = \frac{11}{3}$ .

We need $P(T < 4) = P(T = 0) + P(T = 1) + P(T = 2) + P(T = 3)$ .

Working

With $\lambda = \frac{11}{3}$ ,

P(T < 4) = e^{-\frac{11}{3}}\left(1 + \frac{11}{3} + \frac{\left(\frac{11}{3}\right)^2}{2!} + \frac{\left(\frac{11}{3}\right)^3}{3!}\right)

= 0.501 \quad (3\text{ s.f.})

Answer

0.501

Final answer

0.501

Detailed explanation

Walkthrough

There are two independent samples, so first approximate each sample count by a Poisson distribution. For the first sample, $\lambda_1 = 200 \times \frac{1}{300} = \frac{2}{3}$ . For the second sample, $\lambda_2 = 150 \times 0.02 = 3$ .

Because the samples are independent, the total number of white flowers is the sum of two independent Poisson variables. A key property is that such a sum is also Poisson, with mean equal to the sum of the means: $\lambda = \frac{2}{3} + 3 = \frac{11}{3}$ .

The event "total less than 4" means the total can be 0, 1, 2 or 3. Use the Poisson formula for each of these values and add them:

P(T < 4) = e^{-\frac{11}{3}}\left(1 + \frac{11}{3} + \frac{\left(\frac{11}{3}\right)^2}{2!} + \frac{\left(\frac{11}{3}\right)^3}{3!}\right)

This evaluates to $0.501$ to 3 significant figures.

Key Takeaways

The sum of independent Poisson random variables is Poisson. When approximating two binomial counts by Poisson distributions, add their means before calculating probabilities. "Less than 4" is $0,1,2,3$ , not including 4.

Common Mistakes

Using $\lambda = \frac{2}{3} + 3 = \frac{11}{3}$ but then forgetting the factorial denominators in the Poisson terms.
Including $P(T = 4)$ when the question says "less than 4" (or, equivalently, using $\le 3$ correctly).
Multiplying the Poisson probability by an extra factor. The mark scheme states the expression must not be multiplied by any additional values.

Things to Be Careful About

Check both approximations: $n = 150 > 50$ and $np = 3 < 5$ for the second sample; the first sample was already justified in part (b).
Keep $\lambda = \frac{11}{3}$ in the exponent and in the terms to avoid rounding errors.
Give the final answer to 3 significant figures.

Techniques used

approximate each binomial sample count by a Poisson distributionadd the means of independent Poisson variablescalculate the cumulative Poisson probability for 0, 1, 2 and 3

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