The binomial distribution - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 19 exam-style questions on The binomial distribution for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

An online shop states that 85% of its parcels are delivered on the day after they are ordered. Mo orders 20 parcels from the shop on different days. The random variable X is the number of these parcels that are delivered on the day after they are ordered.

State a suitable distribution for X, and use it to find
the probability that exactly 18 of the parcels are delivered on the day after they are ordered,
the probability that fewer than 16 of the parcels are delivered on the day after they are ordered.

Give your answers to 3 significant figures.

Model answer

\(X \sim B(20, 0.85)\)
\(P(X = 18) = \dbinom{20}{18}(0.85)^{18}(0.15)^2 = 0.229\) (3 s.f.)
\(P(X < 16) = P(X \leqslant 15) = 0.170\) (3 s.f.)

Explanation

There is a fixed number of parcels, each is delivered on the next day or not, and the probability is taken to be the same 0.85 for each parcel, independently, so \(X \sim B(20, 0.85)\). A calculator gives the individual probability \(P(X = 18)\) and the cumulative probability \(P(X \leqslant 15)\), since "fewer than 16" means 15 or fewer. A common error is to use \(P(X \leqslant 16)\), which includes 16.

Mark scheme

1 mark
M1 \(X \sim B(20, 0.85)\) (allow \(B(20, 0.15)\) for the number that are not delivered on the next day, used correctly); or correct final answers of awrt 0.229 and awrt 0.170 with no working
An answer of awrt 0.229 and awrt 0.170 with no working scores 3 marks (one correct answer implies the M1). "Calculator speak" such as "n = 20, p = 0.85" instead of B(20, 0.85) is M0, but the M1 may still be implied by a correct answer.
1 mark
A1 \(P(X = 18) =\) awrt 0.229, i.e. [0.2285, 0.2295)
1 mark
A1 \(P(X < 16) = P(X \leqslant 15) =\) awrt 0.170, i.e. [0.1695, 0.1705)
Depends on the M1. A common error is \(P(X \leqslant 16)\) = 0.352: A0. Allow 0.17 (the 3 s.f. value 0.170 with its final zero dropped).

Question 2

4 marks

In a health screening programme, each person screened is referred for further tests with probability 0.03, independently of other people.

Find the smallest number of people who must be screened for the probability that at least one of them is referred to be greater than 0.95

Model answer

\(P(\text{at least one referred}) = 1 - 0.97^n\)
\(1 - 0.97^n > 0.95\)
\(0.97^n < 0.05\)
\(n \ln 0.97 < \ln 0.05\)
\(n > \dfrac{\ln 0.05}{\ln 0.97} = 98.35\) (the inequality reverses because \(\ln 0.97 < 0\))
Check: \(1 - 0.97^{98} = 0.9495\) and \(1 - 0.97^{99} = 0.9510\)
The smallest number is 99

Explanation

The probability that nobody is referred is \(0.97^n\), so "at least one referred" is \(1 - 0.97^n\), and the condition \(0.97^n < 0.05\) is solved with logarithms or by trying values. Because \(\ln 0.97\) is negative, dividing by it reverses the inequality, giving \(n > 98.35\), so 99 people are needed. Common errors are rounding 98.35 down to 98 and forgetting to reverse the inequality.

Mark scheme

1 mark
M1 \(1 - 0.97^n > 0.95\) (or \(0.97^n < 0.05\)). An answer of 99 with no working scores 2 marks
An answer of 99 with no working scores 2 marks. \(1 - 0.03^n\) or \(0.03n\) scores M0.
1 mark
M1 (dM1) Solves using logarithms, \(n > \dfrac{\ln 0.05}{\ln 0.97}\), or by trials that include \(n = 98\) and \(n = 99\)
Depends on the first M mark.
1 mark
A1 \(n > 98.35\) (awrt 98.4, i.e. [98.35, 98.45)), or \(0.97^{98} = 0.0505\) and \(0.97^{99} = 0.0490\) (values either side of 0.05)
Accept \(1 - 0.97^{98} = 0.9495\) and \(1 - 0.97^{99} = 0.9510\).
1 mark
A1 \(n = 99\)
Depends on both M marks. 98 or 98.35 as the final answer scores A0.

Question 3

5 marks

At a cinema, 22% of customers buy a large popcorn. At one screening there are 30 customers, who decide independently of each other whether to buy a large popcorn.

Find the probability that more than 8 of these customers buy a large popcorn. Give your answer to 3 significant figures.

Before the screening, the kiosk prepares n large popcorns. Find the smallest value of n for which the probability that every customer who wants a large popcorn can have one is at least 0.95

Model answer

Let X be the number of customers who buy a large popcorn: \(X \sim B(30, 0.22)\)
\(P(X > 8) = 1 - P(X \leqslant 8) = 1 - 0.80246 = 0.198\) (3 s.f.)
Everyone who wants one gets one if \(X \leqslant n\), so we need \(P(X \leqslant n) \geqslant 0.95\)
\(P(X \leqslant 9) = 0.8961 < 0.95\) and \(P(X \leqslant 10) = 0.9515 \geqslant 0.95\)
So the smallest value is \(n = 10\)

Explanation

"More than 8" is \(1 - P(X \leqslant 8)\). Every customer can have a popcorn when the number who want one is at most n, so the condition is \(P(X \leqslant n) \geqslant 0.95\); the cumulative probabilities 0.8961 for 9 and 0.9515 for 10 show that 10 is the smallest such n. Showing the values on both sides of 0.95 is what justifies the answer; a common error in the first part is to use \(P(X \leqslant 9)\).

Mark scheme

1 mark
M1 Uses \(X \sim B(30, 0.22)\) with \(P(X > 8) = 1 - P(X \leqslant 8)\); or correct final answers of awrt 0.198 and \(n = 10\) with no working
An answer of awrt 0.198 and \(n = 10\) with no working scores 3 marks (2 for 0.198 and the final A1 for n = 10).
1 mark
A1 awrt 0.198, i.e. [0.1975, 0.1985)
\(1 - P(X \leqslant 9)\) (0.104) scores A0.
1 mark
M1 Interprets the condition as \(P(X \leqslant n) \geqslant 0.95\)
0.95 is the exact value from the question. Accept the condition in words, e.g. "at most n customers want one, with probability at least 0.95".
1 mark
M1 Finds cumulative probabilities either side: \(P(X \leqslant 9) =\) awrt 0.896 (i.e. [0.8955, 0.8965)) and \(P(X \leqslant 10) =\) awrt 0.952 (i.e. [0.9515, 0.9525))
Allow 0.8961 and 0.9515 (4 d.p.).
1 mark
A1 \(n = 10\)
Needs both M marks for this part. \(n = 11\) scores A0.

Question 4

7 marks

The discrete random variable X has probability function
\[P(X = x) = \dfrac{k}{x(x + 1)}, \quad x = 1, 2, 3, 4, 5\]
where k is a constant.

Show that \(k = \dfrac{6}{5}\)

Find the exact value of \(P(X \geqslant 3)\).

Find the probability that X is even, given that \(X \geqslant 2\)

Model answer

\(\sum P(X = x) = k\left(\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + \dfrac{1}{30}\right) = k \times \dfrac{30 + 10 + 5 + 3 + 2}{60} = \dfrac{5}{6}k\)
\(\dfrac{5}{6}k = 1\), so \(k = \dfrac{6}{5}\)
\(P(X \geqslant 3) = \dfrac{6}{5}\left(\dfrac{1}{12} + \dfrac{1}{20} + \dfrac{1}{30}\right) = \dfrac{6}{5} \times \dfrac{10}{60} = \dfrac{1}{5}\)
\(P(X = 2) + P(X = 4) = \dfrac{6}{5}\left(\dfrac{1}{6} + \dfrac{1}{20}\right) = \dfrac{6}{5} \times \dfrac{13}{60} = \dfrac{13}{50}\)
\(P(X \geqslant 2) = 1 - \dfrac{6}{5} \times \dfrac{1}{2} = \dfrac{2}{5}\)
\(P(X \text{ even} \mid X \geqslant 2) = \dfrac{13}{50} \div \dfrac{2}{5} = \dfrac{13}{20}\)

Explanation

The probabilities must add to 1, and with a common denominator of 60 the five fractions add to \(\dfrac{5}{6}\), giving \(k = \dfrac{6}{5}\). \(P(X \geqslant 3)\) adds the last three probabilities (or subtracts the first two from 1). For the conditional probability, the even values that are at least 2 are 2 and 4, and their total is divided by \(P(X \geqslant 2) = \dfrac{2}{5}\); a common error is to forget this division.

Mark scheme

1 mark
M1 Uses \(\sum P(X = x) = 1\) with all five terms: \(k\left(\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + \dfrac{1}{30}\right) = 1\). An answer with no working scores 0 marks for the show that
An answer of \(\dfrac{1}{5}\) and 0.65 with no working scores 5 marks (2 for \(\dfrac{1}{5}\) and 3 for 0.65); the value of k must be shown. Allow the sum written with a common denominator, e.g. \(\dfrac{30 + 10 + 5 + 3 + 2}{60}k = 1\).
1 mark
A1* \(\dfrac{5}{6}k = 1\), so \(k = \dfrac{6}{5}\) (cso)
The sum \(\dfrac{5}{6}\) (or \(\dfrac{50}{60}\)) must be seen. Verification (substituting \(k = \dfrac{6}{5}\) and showing the probabilities add to 1) is acceptable if every probability is shown and a conclusion is given.
1 mark
M1 \(P(X \geqslant 3) = \dfrac{6}{5}\left(\dfrac{1}{12} + \dfrac{1}{20} + \dfrac{1}{30}\right)\) or \(1 - \dfrac{6}{5}\left(\dfrac{1}{2} + \dfrac{1}{6}\right)\)
1 mark
A1 \(\dfrac{1}{5}\)
Accept 0.2 (exact).
1 mark
M1 \(P(X = 2) + P(X = 4) = \dfrac{6}{5}\left(\dfrac{1}{6} + \dfrac{1}{20}\right)\) (\(= \dfrac{13}{50}\))
1 mark
M1 (dM1) Divides by \(P(X \geqslant 2) = 1 - \dfrac{6}{5} \times \dfrac{1}{2}\) (\(= \dfrac{2}{5}\))
Depends on the previous M mark. Dividing by 1 or by \(P(X \geqslant 3)\) scores M0.
1 mark
A1 \(\dfrac{13}{20}\) (0.65)
Exact.

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