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Discrete random variables given by a table or a formula, and the binomial distribution \(B(n, p)\) as a model: its conditions, individual and cumulative probabilities from your calculator, and criticising the model in context.
Binomial conditions: fixed \(n\), two outcomes, independent trials, constant \(p\)
\(P(X \ge r) = 1 - P(X \le r - 1)\)
\(P(X \gt r) = 1 - P(X \le r)\)
Binomial mean \(np\), variance \(np(1 - p)\)
Notes
Discrete random variables
diagram
A discrete random variable \(X\) takes separate values (often counts), each with a probability.
Its probability distribution can be a table or a formula, e.g. \(P(X = x) = kx\) for \(x = 1, 2, 3, 4\). The probabilities add up to 1: \(\Sigma P(X = x) = 1\)
P(X = x) = kx with k = 1/10: the probabilities add up to 1
The cumulative probability \(P(X \le x)\) is the sum of the probabilities up to and including \(x\)
In A level Mathematics you are not asked to calculate the mean or variance of a general discrete random variable.
The binomial distribution
\(X \sim B(n, p)\) models the number of successes in \(n\) trials when: there is a fixed number of trials; each trial has two outcomes (success or failure); the trials are independent; the probability of success, \(p\), is the same for every trial.
X ~ B(15, 0.2): P(X > 4) is everything from 5 upwards
\(P(a \lt X \le b) = P(X \le b) - P(X \le a)\) and \(P(a \le X \le b) = P(X \le b) - P(X \le a - 1)\)
Criticising a binomial model
Independence can fail when trials influence each other, e.g. illness spreading within a class.
\(p\) may not be constant, e.g. when conditions change, or when sampling without replacement from a small population. When sampling from a large population, \(p\) can be treated as constant.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find an unknown constant and probabilities
2 to 4 marksGrade E
Write out \(P(X = x)\) for every value in terms of \(k\)
Add them and set the total equal to 1.
Add the probabilities of the values you need.
Example. The discrete random variable \(X\) has probability distribution \(P(X = x) = k(x + 1)\) for \(x = 1, 2, 3, 4\). (a) Find the value of \(k\). (b) Find \(P(X \ge 3)\)
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(a) \(k(2 + 3 + 4 + 5) = 1\) (M1), so \(14k = 1\) and \(k = \dfrac{1}{14}\)
(A1)
(b) \(\dfrac{4 + 5}{14} = \dfrac{9}{14}\)
(B1)
Model with \(B(n, p)\) and find probabilities
6 to 8 marksGrade C
Define \(X\) and write its distribution, e.g. \(X \sim B(15, 0.2)\)
State assumptions in context: independence and a constant probability.
Rewrite each inequality in terms of \(P(X = r)\) or \(P(X \le r)\)
Use your calculator and give answers to 4 decimal places.
Example. 20% of seeds of a certain type fail to germinate. A gardener plants 15 of these seeds, and \(X\) is the number of seeds that fail to germinate. (a) State two assumptions needed to model \(X\) by \(B(15, 0.2)\). (b) Find \(P(X = 3)\). (c) Find \(P(X \gt 4)\). (d) Find \(P(2 \le X \le 5)\)
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(a) The seeds germinate or fail independently of each other (B1); each seed has the same probability, 0.2, of failing
Pick the condition most likely to fail in this context.
Explain why, referring to the situation.
Example. On a typical day, 10% of the students at a school are absent. A teacher models the number of students absent from a class of 30 on a given day by \(B(30, 0.1)\). Give two reasons why this model may not be suitable.
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Absences may not be independent: illnesses such as colds spread between students in the same class
(B1)
The probability of being absent may not be the same for every student or every day, e.g. there may be more absences in winter
(B1)
Find \(n\) from a probability condition
3 to 4 marksGrade A
Write the condition, often \(1 - P(X = 0) \gt\) a given value.
Rearrange to \((1 - p)^n \lt\) a value.
Take logs: the inequality reverses when you divide by the log of a number less than 1.
Round up to the next whole number and check.
Example. Each component made by a machine is faulty with probability 0.04, independently of other components. Find the smallest number of components that must be checked for the probability of finding at least one faulty component to be greater than 0.9.
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\(1 - 0.96^n \gt 0.9\) (M1), so \(0.96^n \lt 0.1\)
\(n \gt \dfrac{\log 0.1}{\log 0.96} = 56.4\)
(M1)
\(n = 57\)
(A1)
Don’t lose marks
Using \(P(X \le 4)\) for ‘fewer than 4’ (it should be \(P(X \le 3)\)).
Using \(1 - P(X \le 5)\) for \(P(X \ge 5)\) (it should be \(1 - P(X \le 4)\)).
Stating the binomial conditions without context.
Forgetting to reverse the inequality when dividing by the logarithm of a number less than 1.
More tips
Memory tricks
Words to inequalities: ‘at least 3’ is \(X \ge 3\); ‘more than 3’ is \(X \ge 4\); ‘fewer than 3’ is \(X \le 2\); ‘at most 3’ is \(X \le 3\)
Many calculators' binomial CD only gives \(P(X \le r)\), so rewrite everything in that form first.
Swapping to failures: if \(X \sim B(n, p)\) counts successes, the number of failures, \(n - X\), is \(B(n, 1 - p)\)
Exam technique
Write the distribution and the probability statement, e.g. \(X \sim B(15, 0.2)\) and \(P(X \gt 4) = 1 - P(X \le 4)\), before the answer: they earn method marks.
Give probabilities to 4 decimal places (or 3 significant figures or better) unless told otherwise.
The two assumptions usually wanted are independence and a constant probability, both in context.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Use the fact that probabilities sum to 1Solve \(\Sigma P(X = x) = 1\) to find an unknown such as \(k\)
Grade D
Calculate a binomial probability for one valueUse \(P(X = r) = \binom{n}{r}p^r(1 - p)^{n - r}\) or your calculator's binomial PD.
Grade C
Use cumulative binomial probabilities with inequalitiesRewrite ‘more than’, ‘at least’ and ‘between’ in terms of \(P(X \le r)\)
Grade C
State binomial conditions and assumptions in contextFixed number of trials, two outcomes, independent trials, constant probability of success.
Grade B
Criticise a binomial model in contextExplain why independence or a constant probability may not hold.
Grade A
Find \(n\) or \(p\) from a probability conditionE.g. solve \(1 - (1 - p)^n \gt 0.9\) using logarithms.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A student writes down the table below as the probability distribution of a discrete random variable \(X\).
[object Object]
(a) Explain why the table cannot be a probability distribution.[1]
(b) The student’s value for \(P(X = 2)\) is wrong, but all the other values are correct. Find the correct value of \(P(X = 2)\).[1]
(c) Using the corrected distribution, find \(P(X \gt 2)\).[1]
(d) Using the corrected distribution, find \(P(2 \le X \lt 5)\).[2]
Show the answer and mark scheme
(a)Answer: The probabilities add up to 1.05, not 1.
B1 for the sum of the probabilities is 1.05 (not 1)
Worked solution: \(0.1 + 0.35 + 0.25 + 0.2 + 0.15 = 1.05\), but the probabilities of all the values of a random variable must add up to 1.
M1 for \(P(X = 2) + P(X = 3) + P(X = 4)\) with their \(P(X = 2)\)
A1 for 0.75
Worked solution: \(P(2 \le X \lt 5) = 0.3 + 0.25 + 0.2 = 0.75\).
Question 2Medium2 marks
An inspector counts the faulty components in a box of 50 taken from one production run. Faults usually occur in runs when the machine is badly adjusted. A binomial distribution is suggested as a model.
(a) Identify the condition for a binomial model that is not satisfied, and explain why.[1]
(b) Suggest how the data collection could be changed so that a binomial model would be suitable.[1]
Show the answer and mark scheme
(a)Answer: The trials must be independent; the description shows that this is not the case here.
B1 for: the trials must be independent, with a reference to the context
(b)Answer: Take the 50 components at random from several different production runs.
B1 for a sensible change, e.g. take the 50 components at random from several different production runs.
Question 3Hard8 marks
The number of spoonfuls of sugar, \(X\), that a randomly chosen customer at a café adds to a cup of tea has the probability distribution shown in the table, where \(a\) and \(b\) are constants. Given that a customer adds sugar, the probability that they add exactly 2 spoonfuls is \(\frac{1}{3}\).
[object Object]
(a) Find the value of \(b\).[2]
(b) Find the value of \(a\).[1]
(c) Two customers are chosen at random. Find the probability that they add a total of 3 spoonfuls of sugar.[3]
(d) Given that the two customers add a total of 3 spoonfuls, find the probability that one of them adds no sugar.[2]
Show the answer and mark scheme
(a)Answer: \(b = 0.2\)
M1 for \(\dfrac{b}{0.3 + b + 0.1} = \dfrac{1}{3}\)
A1 for \(b = 0.2\)
Worked solution: \(P(X = 2 \mid X \ge 1) = \dfrac{b}{0.4 + b} = \dfrac{1}{3}\), so \(3b = 0.4 + b\) and \(b = 0.2\).
Worked solution: One customer adds no sugar only in the cases (0, 3) and (3, 0), with probability 0.08. So the probability is \(\dfrac{0.08}{0.2} = 0.4\).
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