Discrete random variables

• Table sums to 1; subtract the square of the mean; name the model and its conditions in context before any formula.
• \(\sum p = 1\); \(\mathrm{Var}(X) = \sum x^2p - \{\mathrm{E}(X)\}^2\); fixed \(n\) is binomial, until the first success is geometric; \(\mathrm{P}(X > r) = (1-p)^r\).

• 5.4.1 I can draw up — draw up a probability distribution table relating to a given situation involving a discrete random variable X, and calculate E(X) and Var(X)
• 5.4.2 I can use — use formulae for probabilities for the binomial and geometric distributions, and recognise practical situations where these distributions are suitable models
• 5.4.3 I can use — use formulae for the expectation and variance of the binomial distribution and for the expectation of the geometric distribution

Accuracy for this chapter. Leave a probability exact when it is exact (\(\frac{125}{1296}\), \(\frac{45}{112}\)); otherwise give 3 significant figures. Keep each binomial term to at least 6 decimal places before adding terms, and round once at the end: rounding the terms first can move the third figure. For a least-\(n\) question, quote the values on both sides of the boundary (\(0.8^{10} = 0.1074\), \(0.8^{11} = 0.0859\)) so the reader sees why \(n = 11\) is the least. Write the formula with the numbers substituted before the answer: an unsupported calculator value earns nothing.

• “\(\mathrm{Var}(X) = \sum x^2p = 3.7\).” Correct \(\sum x^2p\) is \(\mathrm{E}(X^2)\), the mean of the squares. The variance is \(\mathrm{Var}(X) = \sum x^2p - \{\mathrm{E}(X)\}^2\): subtract the square of the mean. With \(\mathrm{E}(X) = 1.7\), \(\mathrm{Var}(X) = 3.7 - 2.89 = 0.81\).
• “For a geometric variable, \(\mathrm{P}(X = 0) = p\), and its variance is…” Correct The geometric starts at 1: \(X\) is the number of the trial on which the first success happens, so \(X = 1, 2, 3, \ldots\) and \(\mathrm{P}(X = 1) = p\). And this syllabus gives the geometric no variance: only its mean \(1/p\) is used.
• “Two draws without replacement from 5 red and 3 blue, so \(X \sim \mathrm{B}(2, 5/8)\).” Correct Without replacement the probability changes after the first draw and the draws are not independent, so the model is not binomial. Draw up the distribution from a tree.
• “\(\mathrm{P}(X = 3) = 0.3^3 \times 0.7^7\) for \(\mathrm{B}(10, 0.3)\).” Correct That is the probability of one sequence with three successes. There are \(\binom{10}{3} = 120\) such sequences: \(120 \times 0.3^3 \times 0.7^7 = 0.267\).
• “At least 3 is \(1 - \mathrm{P}(X \le 3)\); more than 3 is \(X \ge 3\).” Correct “At least 3” is \(X \ge 3\), whose complement is \(X \le 2\). “More than 3” is \(X \ge 4\). Write the inequality before the arithmetic.
• “\(\mathrm{P}(X > 4) = 1 - (1-p)^4\) for a geometric variable.” Correct \(X > 4\) means the first four trials all fail: \(\mathrm{P}(X > 4) = (1-p)^4\). The complement \(1 - (1-p)^4\) is \(\mathrm{P}(X \le 4)\).
• A distribution table whose probabilities sum to \(0.9\), used anyway. Repair Every distribution sums to 1. Check it before calculating anything; a sum of \(0.9\) means a value of \(X\) or a branch of the tree is missing.
• “\(\mathrm{Var}(X) = \sum x^2p = 3.7\).” Repair That is \(\mathrm{E}(X^2)\). Subtract \(\{\mathrm{E}(X)\}^2\): \(3.7 - 1.7^2 = 3.7 - 2.89 = 0.81\).
• “\(\mathrm{Var}(X) = \sum x^2p - \mathrm{E}(X) = 3.7 - 1.7 = 2\).” Repair Subtract the square of the mean: \(3.7 - 2.89 = 0.81\).
• “\(X\) = the number of red balls in two draws without replacement, so \(X \sim \mathrm{B}(2, \frac{5}{8})\).” Repair Without replacement \(p\) changes after the first draw and the draws are not independent. Draw up the distribution from a tree: \(\frac{3}{28}, \frac{15}{28}, \frac{10}{28}\), variance \(\frac{45}{112}\), not \(\frac{15}{32}\).
• “\(\mathrm{P}(X = 3) = 0.3^3 \times 0.7^7 = 0.00222\) for \(\mathrm{B}(10, 0.3)\).” Repair That is one sequence. Multiply by \(\binom{10}{3} = 120\), the number of sequences with three successes: \(0.267\).
• “At least 3” computed as \(1 - \mathrm{P}(X \le 3)\). Repair At least 3 is \(X \ge 3\), whose complement is \(X \le 2\): \(1 - \mathrm{P}(X \le 2)\).
• “More than 3” taken as \(X \ge 3\). Repair For a whole-number variable, more than 3 means \(X \ge 4\).
• “\(\mathrm{Var}(X) = np\) for the binomial.” Repair \(np\) is the mean; the variance is \(np(1-p)\).
• Geometric: “\(\mathrm{P}(X = 4) = p(1-p)^4\).” Repair The first success on trial 4 is three failures then a success: \(p(1-p)^3\).
• “\(\mathrm{P}(X > 4) = 1 - (1-p)^4\).” Repair \(X > 4\) means the first four trials fail: \((1-p)^4\). The complement \(1 - (1-p)^4\) is \(\mathrm{P}(X \le 4)\).
• A geometric variable listed with the value 0. Repair \(X\) counts the trial of the first success, so \(X = 1, 2, 3, \ldots\) and \(\mathrm{P}(X = 1) = p\).
• A geometric model for “the number of sixes in 10 throws”. Repair A fixed number of trials, counting successes, is binomial: \(\mathrm{B}(10, \frac{1}{6})\). The geometric counts trials up to the first success.
• A variance quoted for \(\mathrm{Geo}(p)\). Repair The syllabus gives only \(\mathrm{E}(X) = \frac{1}{p}\) for the geometric; no variance is asked for, and none should be offered.
• “\(0.8^n < 0.1\) when \(n = 10.3\), so \(n = 10.3\).” Repair \(n\) is a whole number of trials. Check 10 and 11: \(0.8^{10} = 0.1074\) is too big, \(0.8^{11} = 0.0859\) is small enough, so the least \(n\) is 11.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 27 uses the binomial mean and variance again: when you reach it, re-answer worked example 5 from memory, conditions included. If you go on to Paper 6, its chapters start from the distribution table and \(\sum xp\) of section A, so redo drill 3 of the E and Var drill before you begin. Recalling a method inside a new problem is worth more than another pass over this chapter on its own.

• Chapter 26 · Probability & Statistics 1 · How it is assessed
• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Probability & Statistics 1 content, examined in Paper 5. Paper 5 (Probability & Statistics 1) is 40% of an AS Level that includes it and 20% of the A Level, for which it is compulsory. Its questions use no algebraic methods beyond the Paper 1 content, and it is the foundation for Paper 6. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A situation is described and you draw up the distribution of \(X\), often from a tree or a sample space, sometimes showing that one printed probability is correct first, then find \(\mathrm{E}(X)\) and \(\mathrm{Var}(X)\). Or a context is described, you state the model and its conditions in that context, and the parts ask for single probabilities, a cumulative probability, the mean and variance, or a least \(n\).
• MF19 prints \(\mathrm{E}(X) = \sum xp\), \(\mathrm{Var}(X) = \sum x^2p - \{\mathrm{E}(X)\}^2\), the binomial and geometric formulae, \(np\), \(np(1-p)\) and \(1/p\). The conditions for each model, \(\sum p = 1\) and the geometric tail results \(\mathrm{P}(X > r) = (1-p)^r\) and \(\mathrm{P}(X \le r) = 1 - (1-p)^r\) must be known. There are no binomial tables (MF19 card).
• Probabilities exact where exact, otherwise 3 significant figures; binomial terms kept to at least 6 decimal places before they are added. The formula with its numbers substituted must be visible before each answer, and an unsupported calculator answer earns nothing. A modelling answer states its conditions in the words of the context.