Probability: Sample Space and Favourable Outcomes
Probability appears in the Algebra paper and is assessed through written sample spaces: you list S as a set, count outcomes, and divide. Answers should be given as fractions between 0 and 1, and the sample space is expected in the working.
The basic rule
For equally likely outcomes, P(A) = n(A)/n(S), where n(A) is the number of outcomes favourable to event A and n(S) the number in the sample space. Every probability satisfies 0 ≤ P(A) ≤ 1, with P(S) = 1 and P(impossible event) = 0.
Complement
P(A') = 1 − P(A). For questions saying "at least one", count the complement instead: P(at least one head in two tosses) = 1 − P(TT) = 1 − 1/4 = 3/4.
Standard experiments to memorise
One die: n(S) = 6. Two dice: n(S) = 36, and outcomes are ordered pairs, so (2, 5) and (5, 2) are two different outcomes. Two coins: n(S) = 4, namely HH, HT, TH, TT. A pack of cards: 52 cards, four suits of 13, twelve face cards (J, Q, K in each suit), and four aces, which are not face cards.
Worked example
A card is drawn at random from a well-shuffled pack of 52. Find the probability that it is a king or a heart. P(king) = 4/52, P(heart) = 13/52, and the king of hearts has been counted twice, so subtract 1/52. P = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.
Traps that cost marks
Counting a double such as (4, 4) twice, or omitting the doubles when listing sums of two dice, treating HT and TH as one outcome, counting aces as face cards, and writing a probability greater than 1 after adding cases that overlap are the errors that appear most often. Write the sample space out, then count from it rather than from memory.