The random variable $X$ has the distribution $\text{Po}(5)$. (i) Find $\text{P}(X = 2)$. (ii) It is given that $\text{P}(X = n) = \text{P}(X = n + 1)$. Write down an equation in $n$. (iii) Hence or otherwise find the value of $n$.

(i) 0.0842 (3 sf) · (ii) e^{-5} × 5^n / n! = e^{-5} × 5^{n+1} / (n+1)! · (iii) n = 4

Mathematics 9709/72 — May/June 2019

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

questions

marks

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

Q1The Poisson DistributionFree sample

The random variable $X$ has the distribution $\text{Po}(5)$ .

(i)

Find $\text{P}(X = 2)$ .

DifficultyEasy

Worked solution

Approach

Use the Poisson probability formula $\text{P}(X = x) = e^{-\lambda}\frac{\lambda^x}{x!}$ with $\lambda = 5$ and $x = 2$ .

Working

\text{P}(X = 2) = e^{-5}\frac{5^2}{2!} = e^{-5} \times \frac{25}{2}

Answer

\text{P}(X = 2) = \frac{25e^{-5}}{2} \approx 0.0842

Final answer

0.0842 (3 sf)

Detailed explanation

Walkthrough

The random variable $X$ follows a Poisson distribution with mean $\lambda = 5$ . The probability of exactly $x$ events is given by the formula $\text{P}(X = x) = e^{-\lambda}\frac{\lambda^x}{x!}$ . We substitute $\lambda = 5$ and $x = 2$ into this formula. This gives $e^{-5}\frac{5^2}{2!} = e^{-5} \times \frac{25}{2}$ . Evaluating this with a calculator gives approximately $0.0842$ .

Key Takeaways

The Poisson probability formula is the essential tool for this question. It is important to substitute the correct values of $\lambda$ and $x$ and to compute the factorial correctly.

Common Mistakes

Using the wrong value of $\lambda$ (e.g. using $x$ instead of $\lambda$ ).
Forgetting the factorial in the denominator.
Rounding too early, which could give a slightly different final answer.

Things to Be Careful About

The answer should be given to 3 significant figures as required. The exact form $\frac{25e^{-5}}{2}$ is also acceptable.

Techniques used

substitute into Poisson probability formula

(ii)

It is given that $\text{P}(X = n) = \text{P}(X = n + 1)$ .

Write down an equation in $n$ .

DifficultyMedium-Easy

Worked solution

Approach

Write the Poisson probability formula for $x = n$ and $x = n+1$ , then equate them as given.

Working

\text{P}(X = n) = e^{-5}\frac{5^n}{n!}

\text{P}(X = n+1) = e^{-5}\frac{5^{n+1}}{(n+1)!}

Since $\text{P}(X = n) = \text{P}(X = n+1)$ :

e^{-5}\frac{5^n}{n!} = e^{-5}\frac{5^{n+1}}{(n+1)!}

Answer

e^{-5}\frac{5^n}{n!} = e^{-5}\frac{5^{n+1}}{(n+1)!}

Final answer

e^{-5} × 5^n / n! = e^{-5} × 5^{n+1} / (n+1)!

Detailed explanation

Walkthrough

We use the same Poisson formula but with general values $x = n$ and $x = n+1$ . Writing both probabilities explicitly, we get $e^{-5}\frac{5^n}{n!}$ and $e^{-5}\frac{5^{n+1}}{(n+1)!}$ . The condition $\text{P}(X = n) = \text{P}(X = n+1)$ means these two expressions are equal, giving the required equation.

Key Takeaways

The Poisson formula can be applied symbolically with a general index $n$ . Equating two probabilities gives an equation that can be solved for $n$ .

Common Mistakes

Forgetting that both the exponent and the factorial change when going from $n$ to $n+1$ .
Dropping the $e^{-5}$ factor incorrectly (it can be cancelled later, but it should appear in the equation).

Things to Be Careful About

The equation must be written with the correct powers and factorials. The $e^{-5}$ factor is the same on both sides and can be cancelled, but it is fine to leave it in the equation.

Techniques used

write Poisson probability for general indexequate two probabilities

(iii)

Hence or otherwise find the value of $n$ .

DifficultyMedium-Easy

Worked solution

Approach

Cancel the common factor $e^{-5}$ , then simplify the factorial ratio using $(n+1)! = (n+1)n!$ and solve for $n$ .

Working

From part (ii):

e^{-5}\frac{5^n}{n!} = e^{-5}\frac{5^{n+1}}{(n+1)!}

Cancel $e^{-5}$ :

\frac{5^n}{n!} = \frac{5^{n+1}}{(n+1)!}

Using $(n+1)! = (n+1) \cdot n!$ :

\frac{5^n}{n!} = \frac{5 \cdot 5^n}{(n+1) \cdot n!}

Cancel $5^n$ and $n!$ :

1 = \frac{5}{n+1}

n + 1 = 5

n = 4

Answer

n = 4

Final answer

n = 4

Detailed explanation

Walkthrough

Starting from the equation in part (ii), we cancel the common factor $e^{-5}$ from both sides. Then we use the identity $(n+1)! = (n+1) \cdot n!$ to rewrite the right-hand side. After cancelling $5^n$ and $n!$ , we are left with $1 = \frac{5}{n+1}$ . Multiplying both sides by $n+1$ gives $n+1 = 5$ , so $n = 4$ .

Key Takeaways

Factorials grow by multiplying by the next integer: $(n+1)! = (n+1)n!$ . Cancelling common factors simplifies the equation to a linear one.

Common Mistakes

Incorrectly simplifying the factorial ratio, e.g. writing $\frac{(n+1)!}{n!}$ as something other than $n+1$ .
Making a sign or arithmetic error when solving $1 = \frac{5}{n+1}$ .

Things to Be Careful About

When cancelling, make sure both sides have the same factors. The final value of $n$ must be a non-negative integer, which $n = 4$ is.

Techniques used

cancel common factorssimplify factorial ratiosolve linear equation

The rest of this paper

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