This University probability and random variables revision notes is written for pupils aged 18-19. In short, it shows you how to calculate expectation, variance and use binomial, Poisson and normal distributions. Before you start, it helps to revise multivariable calculus and differential equations, because this lesson builds on them.
University probability and random variables revision notes: what you will learn
Probability gives a precise language for uncertainty. This chapter builds from the axioms to conditional probability, random variables and the binomial, Poisson and normal distributions.
Starter activity
Roll two dice twenty times and record whether you get a double. Before you start, predict how many doubles you expect. Compare your result with the rest of the group: why are the results different, and why do the averages get closer to the prediction as more results are pooled? Then ask: if the first die shows a six, does that change the chance of a double?
Probability foundations
Definition: probability space
A probability space consists of a sample space \(\Omega\), a collection of events (subsets of \(\Omega\)) and a probability \(P\) satisfying the Kolmogorov axioms: \(P(A)\ge0\), \(P(\Omega)=1\), and \(P(A\cup B)=P(A)+P(B)\) for disjoint events (extended to countable unions).
Rule: consequences of the axioms
\(P(A^c)=1-P(A)\), and the addition rule is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
Example 1
If \(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\cap B)=0.2\), find \(P(A\cup B)\).
\(0.5+0.4-0.2=0.7\).
Conditional probability, independence and Bayes’ theorem
Definition: conditional probability and independence
\(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\) for \(P(B)>0\). Events are independent if \(P(A\cap B)=P(A)P(B)\), equivalently \(P(A|B)=P(A)\).
Theorem: total probability and Bayes' theorem
For a partition \(\{A_i\}\): \(P(B)=\sum P(B|A_i)P(A_i)\) and \(P(A_i|B)=\dfrac{P(B|A_i)P(A_i)}{P(B)}\).
Method: Bayes problems
- Draw a tree diagram with the prior probabilities on the first branches.
- Add the conditional probabilities on the second branches.
- Find \(P(B)\) by adding the branches that end in \(B\).
- Divide the branch you want by \(P(B)\).
Example 2
A disease affects 1% of people. A test detects it 95% of the time but gives a false positive for 5% of healthy people. Find \(P(D|+)\).
\(P(D|+)=\frac{0.95\times0.01}{0.95\times0.01+0.05\times0.99}=\frac{0.0095}{0.0590}\approx0.161\). A positive test means only a 16% chance of disease, because the disease is rare.
Example 3
With the numbers of Example 1, check independence and find \(P(A|B)\).
\(P(A|B)=\frac{0.2}{0.4}=0.5=P(A)\), and \(0.5\times0.4=0.2\), so \(A\) and \(B\) are independent. For counting, \(\binom{10}3=120\) committees of 3 can be chosen from 10 people.
Random variables, expectation and variance
Definition: expectation and variance
A random variable \(X\) assigns a number to each outcome. For a discrete \(X\) with pmf \(p(x)=P(X=x)\): \(E[X]=\sum xp(x)\) and \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\). For a continuous \(X\) with density \(f\): \(P(a\lt X\lt b)=\int_a^bf\), \(\int f=1\), \(E[X]=\int xf(x)\,dx\).
Rule: linearity
\(E[aX+b]=aE[X]+b\) and \(\operatorname{Var}(aX+b)=a^2\operatorname{Var}X\). Expectation is always additive; variance is additive when the variables are independent.
Example 4
For \(P(0)=0.2\), \(P(1)=0.5\), \(P(2)=0.3\), find \(E[X]\), \(\operatorname{Var}X\), \(E[3X+2]\) and \(\operatorname{Var}(3X+2)\).
\(E[X]=1.1\), \(E[X^2]=1.7\), \(\operatorname{Var}X=1.7-1.21=0.49\). Then \(E[3X+2]=5.3\) and \(\operatorname{Var}(3X+2)=9\times0.49=4.41\).
Example 5
A density is \(f(x)=kx\) on \([0,2]\). Find \(k\), \(E[X]\) and \(P(X<1)\).
\(\int_0^2kx\,dx=2k=1\), so \(k=\frac12\). \(E[X]=\int_0^2\frac{x^2}2dx=\frac43\) and \(P(X<1)=\int_0^1\frac x2dx=\frac14\).
The binomial distribution
Rule: binomial distribution
If \(X\) counts successes in \(n\) independent trials with success probability \(p\), then \(X\sim B(n,p)\) with \(P(X=r)=\binom nrp^r(1-p)^{n-r}\), \(E[X]=np\) and \(\operatorname{Var}X=np(1-p)\).
Example 6
For \(X\sim B(10,0.3)\) find \(P(X=3)\). What is the chance of at least one success in 5 trials with \(p=0.2\)?
\(P(X=3)=120(0.3)^3(0.7)^7\approx0.2668\). And \(1-0.8^5=0.6723\).
The Poisson distribution
Rule: Poisson distribution
For rare events at a constant average rate \(\lambda\) per interval, \(X\sim\text{Po}(\lambda)\) with \(P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!}\) and \(E[X]=\operatorname{Var}X=\lambda\). It approximates \(B(n,p)\) when \(n\) is large and \(p\) small, with \(\lambda=np\).
Example 7
For \(\lambda=3\) find \(P(X=2)\) and \(P(X=0)\).
\(\frac{9e^{-3}}2\approx0.2240\) and \(e^{-3}\approx0.0498\). For \(n=200\), \(p=0.01\), \(\lambda=2\), \(P(X=1)\approx2e^{-2}=0.2707\), matching the binomial value.
The normal distribution
Rule: standardising
\(X\sim N(\mu,\sigma^2)\) has a bell-shaped density symmetric about \(\mu\). Standardise \(Z=\frac{X-\mu}\sigma\sim N(0,1)\) and use \(\Phi(z)=P(Z\le z)\). Key values: \(\Phi(1)=0.8413\), \(\Phi(2)=0.9772\), \(\Phi^{-1}(0.95)=1.645\). Sums of independent normals are normal, with means and variances added.
Example 8
For \(X\sim N(100,15^2)\) find \(P(X>115)\), \(P(85\lt X<130)\) and the 95th percentile.
\(1-\Phi(1)=0.1587\); \(\Phi(2)-\Phi(-1)=0.8186\); \(100+1.645\times15\approx124.7\). If \(X\sim N(5,4)\) and \(Y\sim N(3,9)\) are independent then \(X+Y\sim N(8,13)\).
Watch out
- Confusing independent with mutually exclusive events.
- Forgetting that variance involves the square of constants.
- Adding standard deviations instead of variances.
- Neglecting the base rate in Bayes problems.
Using the distributions
Key fact: approximating one distribution by another
Exact binomial calculations become heavy when \(n\) is large. If \(n\) is large and \(p\) small (roughly \(n\ge50\) and \(np<5\)), \(B(n,p)\) is close to \(\text{Po}(np)\), because rare events in many trials behave like random arrivals. If \(np\) and \(n(1-p)\) are both larger than about 5, \(B(n,p)\) is close to \(N(np,\,np(1-p))\), and a Poisson distribution with large \(\lambda\) is close to \(N(\lambda,\lambda)\). When a discrete variable is approximated by a continuous one, use a continuity correction: \(P(X\le k)\) becomes \(P(Y\lt k+0.5)\) and \(P(X\ge k)\) becomes \(P(Y>k-0.5)\).
Method: normal probabilities and percentiles
- Write the distribution clearly, for example \(X\sim N(\mu,\sigma^2)\), and note whether the second parameter is the variance or the standard deviation.
- Standardise: \(Z=\dfrac{X-\mu}\sigma\sim N(0,1)\).
- Sketch the bell curve and shade the region you need; use symmetry \(\Phi(-z)=1-\Phi(z)\).
- For a percentile, find \(z\) with \(\Phi(z)=\) the required proportion, then return to \(x=\mu+z\sigma\).
- Check that the answer is sensible: probabilities between 0 and 1, percentiles on the correct side of the mean.
Worked example 9: Bayes with a table of natural frequencies
5% of emails are spam. A filter flags 98% of spam and 1% of genuine emails. What proportion of flagged emails are spam?
Think of 10 000 emails: 500 are spam, of which 490 are flagged; 9 500 are genuine, of which 95 are flagged. Altogether 585 emails are flagged, and 490 of them are spam. So \(P(\text{spam}\mid\text{flagged})=\dfrac{490}{585}\approx0.838\). The same answer comes from Bayes' theorem: \(\dfrac{0.05\times0.98}{0.05\times0.98+0.95\times0.01}=\dfrac{0.049}{0.0585}\). Natural frequencies make it easy to see that the false alarms come from the large genuine group.
Worked example 10: Variance of a sum
\(X\) and \(Y\) are independent with \(E[X]=3\), \(\operatorname{Var}(X)=4\), \(E[Y]=5\), \(\operatorname{Var}(Y)=9\). Find the mean and variance of \(2X-Y+1\).
By linearity, \(E[2X-Y+1]=2\times3-5+1=2\). For the variance the constant has no effect, and for independent variables \(\operatorname{Var}(aX+bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)\). So \(\operatorname{Var}(2X-Y+1)=4\times4+1\times9=25\) and the standard deviation is 5. Note that the minus sign disappears when squared: subtracting \(Y\) adds its variance.
Worked example 11: A Poisson model in context
Typing errors occur at random at an average rate of 1.5 per page. Find the probability that a 4-page essay has at most 3 errors.
For 4 pages the number of errors is \(\text{Po}(6)\). Then \(P(X\le3)=e^{-6}\left(1+6+\dfrac{36}2+\dfrac{216}6\right)=e^{-6}(1+6+18+36)=61e^{-6}\approx0.151\). The key step is to scale the rate to the interval in the question: the Poisson parameter is always the mean number of events in that interval.
Worked example 12: A normal approximation with continuity correction
A fair coin is tossed 100 times. Estimate the probability of at least 60 heads.
\(X\sim B(100,0.5)\) has mean 50 and variance 25, so \(X\approx N(50,25)\). With the continuity correction, \(P(X\ge60)\approx P(Y>59.5)=P\left(Z>\dfrac{9.5}5\right)=P(Z>1.9)=1-0.9713=0.0287\). The exact binomial value is \(0.0284\), so the approximation is excellent. Without the correction we would get \(P(Z>2)=0.0228\), noticeably worse.
Worked example 13: A continuous random variable
A random variable has density \(f(x)=\tfrac38x^2\) on \([0,2]\). Find its cumulative distribution function and median.
For \(0\le x\le2\), \(F(x)=\int_0^x\tfrac38t^2\,dt=\dfrac{x^3}8\); also \(F(x)=0\) for \(x<0\) and \(F(x)=1\) for \(x>2\). The median \(m\) satisfies \(F(m)=\tfrac12\), so \(m^3=4\) and \(m=\sqrt[3]4\approx1.587\). The mean is \(\int_0^2\tfrac38x^3\,dx=\tfrac32\), a little less than the median, because the density is skewed towards the right end of the interval.
Worked example 14: Expected value as a long-run average
A raffle sells 500 tickets at £1. There is one prize of £100 and four prizes of £20. Find the expected gain of a ticket holder.
Let \(G\) be the gain. A ticket wins £100 with probability \(\frac1{500}\) and £20 with probability \(\frac4{500}\); otherwise it wins nothing. The expected prize is \(\dfrac{100+4\times20}{500}=\dfrac{180}{500}=£0.36\), so the expected gain is \(0.36-1=-£0.64\) per ticket. The organisers keep, on average, 64p from every ticket sold, that is \(500\times0.64=£320\) for the whole raffle. An expected value is not a value you will actually see on one ticket: it is the average gain per ticket over many tickets, which is why it is the right tool for judging whether a game is fair.
Note: modelling assumptions
Every distribution in this chapter comes with assumptions. The binomial model needs a fixed number of independent trials with the same probability of success; the Poisson model needs events that occur singly, independently and at a constant average rate; the normal model is justified for measurements that are the sum of many small independent effects. In applications, always state the assumptions in context and ask whether they are reasonable: calls to a helpline may come in bursts after an advert, and the heights of a mixed group of adults may be better described by two overlapping normal curves. A good habit is to compare the model with data: if the sample mean and variance of a count are very different, a Poisson model is doubtful; if a histogram of measurements is strongly skewed, a normal model is doubtful. Models are simplifications, and their conclusions are only as reliable as their assumptions.
Watch out: typical mistakes
- Confusing \(P(A\mid B)\) with \(P(B\mid A)\), the classic error in medical and legal reasoning.
- Assuming that mutually exclusive events are independent: they are not, unless one has probability 0.
- Writing \(N(\mu,\sigma)\) when the notation means \(N(\mu,\sigma^2)\).
- Forgetting that \(\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)\) for independent variables.
Did you know?
Probability began with letters between Pascal and Fermat in 1654 about a gambling problem. Thomas Bayes' famous theorem was published after his death in 1763, and the Poisson distribution appeared in 1837 in a study of court verdicts. Its most famous early use was by Ladislaus Bortkiewicz in 1898, who showed that the number of Prussian soldiers killed by horse kicks each year followed a Poisson distribution very closely.
Otto's tip
Draw a tree diagram for conditional probability questions.
How to use this University probability and random variables revision notes
First, read the key points so that you know where the University probability and random variables revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the probability and random variables worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the probability and random variables test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this University probability and random variables revision notes gives you everything you need for the chapter.
To sum up, regular practice works better than one long session. For instance, ten minutes a day is often enough. Besides, mistakes are useful, because they show what to revise next. Then, once a method feels easy, try a harder question. As a result, you build confidence step by step.
Similar pages
WorksheetProbability and random variables – university Year 1 maths worksheet with answers to download as PDF
AnswersProbability and random variables – university Year 1 maths worksheet answers to download as PDF
Multiple choice questionsProbability and random variables – university Year 1 maths multiple choice questions to download as PDF
QuizProbability and random variables – university Year 1 maths quiz to download as PDF
TestProbability and random variables – university Year 1 maths test to download as PDF
Test answersProbability and random variables – university Year 1 maths test answers to download as PDF
StartersProbability and random variables – university Year 1 maths starters to download as PDFOther chapters in this level
LessonLogic, sets and proof techniques – university Year 1 maths lesson to download as PDF
LessonFunctions, limits and continuity – university Year 1 maths lesson to download as PDF
LessonDifferentiation and its applications – university Year 1 maths lesson to download as PDF
LessonIntegration and the fundamental theorem of calculus – university Year 1 maths lesson to download as PDF
LessonSequences and series convergence – university Year 1 maths lesson to download as PDF
LessonVectors and systems of linear equations – university Year 1 maths lesson to download as PDF
LessonMatrices and determinants – university Year 1 maths lesson to download as PDF
LessonVector spaces and linear maps – university Year 1 maths lesson to download as PDF