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 (3 sf) · (b) n > 50 and np = 2/3 < 5 (or p = 1/300 < 0.1) · (c) 0.501 (3 sf)

Mathematics 9709/61 — October/November 2020

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

questions

marks

minutes

Topics Linear Combinations of Random Variables · The Poisson Distribution · 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

The number of white flowers has a binomial distribution, $X \sim \text{B}(200, \frac{1}{300})$ . Since $n=200>50$ and $np=\frac{2}{3}<5$ , the Poisson approximation is appropriate, with $\lambda = np = \frac{2}{3}$ . We need $P(X>1)$ , so we calculate $1-P(X=0)-P(X=1)$ .

Working

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

Using the Poisson formula $P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$ :

\begin{aligned} P(X>1) &= 1 - P(X=0) - P(X=1) \\ &= 1 - e^{-2/3}\left(1 + \frac{2}{3}\right) \\ &= 1 - 0.855695... \\ &= 0.144305... \end{aligned}

Answer

P(X>1) = 0.144 \quad (3 \text{ sf})

Final answer

0.144 (3 sf)

Detailed explanation

Walkthrough

We have a fixed number of trials, 200, and each trial has the same small probability of success, $1/300$ . The exact distribution is therefore binomial. For a Poisson approximation to be used, we need $n$ large and $p$ small; here $n=200>50$ and $np=2/3<5$ . The approximating distribution is $\text{Po}(2/3)$ . To find the probability of more than one white flower, we use the complement: $P(X>1)=1-P(X=0)-P(X=1)$ . Using the Poisson formula gives the numerical answer.

Key Takeaways

This question tests the Poisson approximation to the binomial distribution and the use of the complement rule for a tail probability. It also reinforces that the mean of the approximating Poisson distribution is $np$ .

Common Mistakes

Using the binomial distribution directly instead of the Poisson approximation.
Forgetting to subtract $P(X=1)$ as well as $P(X=0)$ when finding $P(X>1)$ .
Giving no working, or rounding before the final step.

Things to Be Careful About

'More than 1' means $X=2,3,4,...$ , so it is $1-P(X=0)-P(X=1)$ , not $1-P(X=1)$ .
The mean is $\lambda=\frac{2}{3}$ , not 200 or 1/300.
Give the final answer to 3 significant figures.

Techniques used

identify the binomial modelapply the Poisson approximation with λ = npcalculate a complement probability P(X > 1)

(b)

Justify the approximating distribution used in part (a).

DifficultyEasy

Worked solution

Approach

The Poisson approximation to the binomial distribution is valid when $n$ is large and $p$ is small. The standard numerical checks are $n>50$ and $np<5$ , or equivalently $n>50$ and $p<0.1$ .

Working

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

(or $n>50$ and $p=\frac{1}{300}<0.1$ ).

Final answer

n > 50 and np = 2/3 < 5 (or p = 1/300 < 0.1)

Detailed explanation

Walkthrough

The binomial distribution can be approximated by a Poisson distribution when the number of trials is large and the probability of success is small. The standard numerical checks are $n>50$ and $np<5$ , or equivalently $n>50$ and $p<0.1$ . Here $n=200>50$ and $np=200\times\frac{1}{300}=\frac{2}{3}<5$ , so the approximation used in part (a) is valid.

Key Takeaways

A justification of the Poisson approximation must quote specific conditions, not just say 'n is large and p is small'. The conditions can be written using $np$ or $p$ .

Common Mistakes

Saying only 'n is large and p is small' without numerical checks.
Forgetting to state $n>50$ .
Using $p$ without relating it to the given probability.

Things to Be Careful About

The mark scheme accepts either $np<5$ or $p<0.1$ , but the condition must be clearly stated.
Use the actual values from the question: $n=200$ , $p=1/300$ .

Techniques used

state the conditions for a Poisson approximationverify 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 $X$ be the number of white flowers in the first sample and $Y$ the number in the second. Each is binomial and is approximated by a Poisson distribution. Since the samples are independent, the total $T=X+Y$ is approximately Poisson with mean equal to the sum of the two means. We need $P(T<4)=P(T=0)+P(T=1)+P(T=2)+P(T=3)$ .

Working

First sample: $X \sim \text{B}(200, \frac{1}{300})$ , so

X \approx \text{Po}\left(200 \times \frac{1}{300}\right) = \text{Po}\left(\frac{2}{3}\right)

Second sample: $Y \sim \text{B}(150, 0.02)$ , so

Y \approx \text{Po}(150 \times 0.02) = \text{Po}(3)

Since the samples are independent,

T = X+Y \approx \text{Po}\left(\frac{2}{3}+3\right) = \text{Po}\left(\frac{11}{3}\right)

Then

\begin{aligned} P(T<4) &= P(T=0)+P(T=1)+P(T=2)+P(T=3) \\ &= e^{-11/3}\left(1+\frac{11}{3}+\frac{(11/3)^2}{2!}+\frac{(11/3)^3}{3!}\right) \\ &= 0.501... \end{aligned}

Answer

P(\text{total white flowers}<4) = 0.501 \quad (3 \text{ sf})

Final answer

0.501 (3 sf)

Detailed explanation

Walkthrough

There are two independent samples, so we approximate each binomial count by a Poisson distribution. For the first sample, $\lambda_1=200\times\frac{1}{300}=\frac{2}{3}$ . For the second sample, $\lambda_2=150\times0.02=3$ . Because the samples are independent, the total $T=X+Y$ is approximately Poisson with mean $\lambda_1+\lambda_2=\frac{11}{3}$ . The event 'less than 4' means $T=0,1,2,3$ , so we add the four Poisson probabilities. The expression is exactly the Poisson cumulative probability with no extra multipliers.

Key Takeaways

This question combines two ideas: the Poisson approximation to the binomial distribution and the fact that the sum of independent Poisson random variables is Poisson with mean equal to the sum of the means.

Common Mistakes

Using the binomial distribution for each sample and trying to combine binomial probabilities.
Forgetting to add the two means, or adding the sample sizes instead.
Including $P(T=4)$ when the question asks for 'less than 4'.
Multiplying the Poisson expression by an extra factor.

Things to Be Careful About

'Less than 4' means $T=0,1,2,3$ only.
The combined mean is $\frac{11}{3}$ , not $\frac{2}{3}$ or 3 alone.
The mark scheme requires the expression to be exactly $e^{-11/3}(1+\frac{11}{3}+\frac{(11/3)^2}{2!}+\frac{(11/3)^3}{3!})$ ; do not multiply by anything else.
Give the final answer to 3 significant figures.

Techniques used

approximate each binomial distribution by a Poisson distributionadd the means of independent Poisson variablescalculate a cumulative Poisson probability P(T < 4)

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