Probability

• Exclusive events add; independent events multiply; the given event is the denominator; and \(\text{P}(A \mid B)\) is not \(\text{P}(B \mid A)\).
• Exclusive: P(A ∩ B) = 0, add. Independent: P(A ∩ B) = P(A) × P(B), multiply, and test it with both numbers. Conditional: divide by the given event.

• 5.3.1 I can evaluate — evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations
• 5.3.2 I can use — use addition and multiplication of probabilities, as appropriate, in simple cases
• 5.3.3 I can understand — understand the meaning of exclusive and independent events, including determination of whether events A and B are independent by comparing the values of P(A ∩ B) and P(A) × P(B)
• 5.3.4 I can calculate — calculate and use conditional probabilities in simple cases

Accuracy for this chapter. Leave an exact probability as a fraction in lowest terms (\(\tfrac{5}{14}\), \(\tfrac{12}{19}\)); where a decimal is wanted or the value is not exact, give 3 significant figures (\(\tfrac{53}{66} = 0.803\), \(\tfrac{12}{19} = 0.632\)), carrying full precision until the end. A probability built from decimal data is usually exact as a decimal (\(0.42\), \(0.19\)): leave it so. A probability has no unit, and it lies between 0 and 1; an answer greater than 1 is a sign that overlapping events were added. Of the chapter 24 counts, \({}^{n}C_{r}\) is printed in MF19, as the binomial coefficient \(\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\) with the binomial series; \({}^{n}P_{r}\) is not, and must be known.

• “A and B are mutually exclusive, so they are independent.” Correct If A and B are exclusive and both have non-zero probability, then \(\text{P}(A \cap B) = 0\) but \(\text{P}(A) \times \text{P}(B) > 0\), so they are not independent: knowing A happened tells you B did not. Exclusive is about whether the events can overlap; independent is about whether one changes the probability of the other (5.3.3).
• “P(boy | plays) and P(plays | boy) are the same thing written two ways.” Correct The event after the bar is the one known to have happened, and its total is the denominator. In the table of 80 students, \(\text{P}(\text{plays} \mid \text{boy}) = \tfrac{18}{40}\) but \(\text{P}(\text{boy} \mid \text{plays}) = \tfrac{18}{42}\): same overlap, different given event (5.3.4).
• “Two dice have 11 possible totals, so P(total 7) = 1/11.” Correct The totals are not equally likely; the 36 ordered pairs are. Six of them total 7, so \(\text{P}(7) = \tfrac{6}{36} = \tfrac{1}{6}\) (5.3.1).
• “P(A or B) = P(A) + P(B), always.” Correct Only when A and B are exclusive. When they can happen together, adding counts the overlap twice: draw a Venn diagram or a two-way table and add the regions (5.3.2).
• “Without replacement, both red is 5/8 × 5/8.” Correct After one red ball is taken the bag holds 4 red among 7, so the second branch is the conditional probability \(\tfrac{4}{7}\): \(\tfrac{5}{8} \times \tfrac{4}{7} = \tfrac{5}{14}\).
• “They look unrelated, so they are independent.” Correct Independence is decided by calculation, not by appearance: find \(\text{P}(A \cap B)\) and \(\text{P}(A) \times \text{P}(B)\) from the data and compare them, stating both numbers.
• “11 possible totals, so P(total 7) = 1/11.” Repair The totals are not equally likely; the 36 ordered pairs are. \(\text{P}(7) = \tfrac{6}{36} = \tfrac{1}{6}\).
• “P(total 8) = 3/36: (2, 6), (3, 5), (4, 4).” Repair \((6, 2)\) and \((5, 3)\) are different outcomes from \((2, 6)\) and \((3, 5)\): five cells, \(\tfrac{5}{36}\).
• “Without replacement, P(both red) = 5/8 × 5/8.” Repair The second draw has 4 red among 7: \(\tfrac{5}{8} \times \tfrac{4}{7} = \tfrac{5}{14}\).
• “P(one of each) = 5/8 × 3/7.” Repair That is red then blue. Add blue then red: \(\tfrac{15}{56} + \tfrac{15}{56} = \tfrac{15}{28}\).
• “Favourable: 5 × 3 = 15 pairs. Total: \({}^{8}P_{2}\) = 56. So 15/56.” Repair Favourable counted as selections, total as arrangements. Count both the same way: \(\tfrac{15}{{}^{8}C_{2}} = \tfrac{15}{28}\).
• “P(A or B) = P(A) + P(B).” (for events that overlap) Repair Add only when the events are exclusive; otherwise fill in a Venn diagram or a two-way table and add its regions.
• “P(at least one six in three throws) = 3 × 1/6 = 1/2.” Repair The three events overlap, so they cannot be added. \(1 - \left(\tfrac{5}{6}\right)^3 = \tfrac{91}{216}\).
• “A and B are exclusive, so they are independent.” Repair Exclusive events with non-zero probabilities are never independent: \(\text{P}(A \cap B) = 0 \ne \text{P}(A)\text{P}(B)\).
• “Independent means they cannot happen together.” Repair That is exclusive. Independent means the occurrence of one does not affect the probability of the other.
• “They look unrelated, so they are independent.” Repair Determine it: compare \(\text{P}(A \cap B)\) with \(\text{P}(A) \times \text{P}(B)\), both calculated, and state both numbers.
• “P(boy | plays) = 18/40.” Repair The condition “plays” is the column of 42: \(\tfrac{18}{42} = \tfrac{3}{7}\). \(\tfrac{18}{40}\) is \(\text{P}(\text{plays} \mid \text{boy})\).
• “P(rain | late) = 0.4.” Repair 0.4 is \(\text{P}(\text{late} \mid \text{rain})\), a branch on the tree. \(\text{P}(\text{rain} \mid \text{late}) = \tfrac{0.12}{0.19} = \tfrac{12}{19}\).
• “P(A | B) = P(A ∩ B) / P(A).” Repair Divide by the probability of the event that is given: \(\text{P}(A \mid B) = \dfrac{\text{P}(A \cap B)}{\text{P}(B)}\).

• 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 26 builds every probability distribution from these rules: when you reach the binomial and geometric distributions, re-answer retrieval question 11 as a binomial probability. Chapter 27 uses the complement in every normal tail: re-answer why P(X > a) = 1 − P(X ≤ a). Chapter 24’s counts return in every probability by combinations here. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Chapter 25 · 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 structured question on this section can turn a context into a tree or a two-way table, ask for a probability read from it, then for a conditional probability “given” one of the events, or ask you to determine whether two events are independent. Selections from a group can combine this section with the counting of chapter 24. The reasoning is in the set-up: which outcomes are equally likely, which branches are conditional, and which event is given.
• MF19 prints no probability rule for this section. The equally-likely definition, the complement, addition for exclusive events, multiplication for independent events and \(\text{P}(A \mid B) = \text{P}(A \cap B) \div \text{P}(B)\) must all be known, with their conditions (MF19 card).
• Leave exact probabilities as fractions in lowest terms (\(\tfrac{12}{19}\)), otherwise 3 significant figures. Write each path as a product before its value and each conditional as a quotient before its value. An independence conclusion states both \(\text{P}(A \cap B)\) and \(\text{P}(A) \times \text{P}(B)\); a bare “independent” is unsupported.