Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 5: Probability & Statistics 1 · official syllabus
DISCRETE RANDOM VARIABLES LAB
Learn it by playing it
Four short missions on expectation, variance, binomial counts and geometric waiting times. Predict the rule, then calculate using the stated probability model.
1234
Mission 1 of 4
Find the long-run mean
X takes values 0, 1 and 2 with probabilities 0.2, 0.5 and 0.3. Set E(X) in tenths: a value 11 means 1.1. Weight each possible value by its probability, rather than averaging the three labels equally.
Must E(X) be a possible observed value of X?
Bar heights show P(X=x), not frequencies. The expectation weights the values 0, 1 and 2 by their respective probabilities.
Your expectation is 5/10. Multiply each value by its own probability.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Build a discrete probability distribution 5.4
A discrete random variable assigns a numerical value to each outcome of an experiment and has a finite or countably infinite set of possible values. Define what X measures before writing its distribution. Different elementary outcomes can give the same value of X, so add their probabilities.
A probability table lists each possible value x and P(X=x). All probabilities must be non-negative and sum to 1. A missing probability can be found from this total, but the result must also be between 0 and 1. Impossible values have probability zero.
Worked example: X takes values 0, 1, 2 with probabilities 0.2, 0.5, 0.3. These probabilities sum to 1. P(X at least 1)=0.5+0.3=0.8, while P(X greater than 1)=0.3. The inequality decides whether the endpoint is included.
Expectation is a weighted mean 5.4
E(X)=sum of xP(X=x), over all possible values. It describes the long-run average under repeated use of the same probability model. It is not necessarily a possible observation, the most likely observation, or a guaranteed outcome.
For X=0,1,2 with probabilities 0.2,0.5,0.3, E(X)=0(0.2)+1(0.5)+2(0.3)=1.1. Averaging 0,1,2 without weights would give 1 and would incorrectly treat these unequal probabilities as equal.
To find E(g(X)), sum g(x)P(X=x). In particular, E(X squared)=sum of x squared times P(X=x). Square the values before weighting them, not the probabilities. For the example E(X squared)=0+0.5+1.2=1.7.
Variance measures spread about the mean 5.4
Var(X)=E(X squared)-[E(X)] squared. Equivalently it is the weighted mean of (x-E(X)) squared. Standard deviation is the positive square root of variance. Variance has squared units; standard deviation has the same units as X.
For the example, Var(X)=1.7-1.1 squared=0.49 and standard deviation=0.7. A direct check gives 0.2(0-1.1)^2+0.5(1-1.1)^2+0.3(2-1.1)^2=0.49. Keep enough precision to avoid making a small variance negative by rounding.
Variance is non-negative. If a calculation gives a negative value, check the table total, squared values, weighted sum and subtraction order. Do not report the negative value or silently replace it by its absolute value.
Recognise and use a binomial model 5.4
X has distribution B(n,p) when X counts successes in a fixed number n of independent trials, each with two outcomes called success and failure, with the same success probability p. Check these assumptions from the situation. Sampling without replacement from a small finite population generally changes p and is not exactly binomial.
For r=0,1,...,n, P(X=r)=nCr p^r(1-p)^(n-r). The combination chooses the success positions. Multiply a single sequence probability by the number of such sequences. For B(4,1/2), P(X=2)=4C2(1/2)^2(1/2)^2=6/16=3/8.
For B(n,p), E(X)=np and Var(X)=np(1-p). Thus for B(4,1/2), the expectation is 2 and variance is 1. A cumulative probability requires a sum of the appropriate terms, or a complement: P(X at least 1)=1-P(X=0)=1-(1-p)^n.
Recognise and use a geometric model 5.4
T has distribution Geo(p) when independent binary trials with constant success probability p continue until the first success, and T counts the trials including the successful trial. The possible values are 1,2,3,... . This trial-count convention matters: some other sources count preceding failures instead.
For r=1,2,3,..., P(T=r)=(1-p)^(r-1)p. There must be r-1 failures followed by one success, in that order. No binomial coefficient is needed. With p=1/4, P(T=3)=(3/4)^2(1/4)=9/64.
E(T)=1/p, so with p=1/4 the expected wait is 4 trials. P(T>r)=(1-p)^r, since the first r trials must all fail. Therefore P(T at most r)=1-(1-p)^r. For p=1/4, P(T>3)=27/64 and P(T at most 3)=37/64.
Choose the variable and model before calculating 5.4
A fixed four-trial success count and a wait for the first success are different variables. For four trials, exactly two successes allows multiple success positions. For first success on trial 3, the success position is fixed and the first two trials must be failures.
Explain why a model is suitable rather than naming it from a keyword. An independence assumption and constant p may be reasonable for repeated randomised trials; changing conditions, dependence or a stopping rule can invalidate a binomial model. A fixed number of trials is not a geometric waiting-time experiment.
Use exact fractions in working when convenient and round only at the requested stage. For a finite distribution, independently check the probability total and expectation's range. For a binomial count, r must be an integer from 0 to n; for the stated geometric convention, r starts at 1.
EduTribe study notes - original material aligned to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026-2027). Automated checks completed; not yet reviewed by an educator.
Free public preview - no sign-in or pass needed. Automated checks completed; not yet reviewed by an educator. Sign in to save your practice progress across sessions.