Rae

HomeSubjectsA-Level H2 Maths › Probability

A-Level Probability

What the A-Level syllabus expects for Probability, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to find, determine, calculate, name. About 7% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (3 marks)

From the bag of 4 red and 6 blue balls, drawn one at a time without replacement until empty, find the probability that the first five draws contain exactly 2 blue balls, given they contain at least 3 red balls.

Show the worked answer

Bag has 4 red and 6 blue (10 total). Consider the first five draws; number of red among them can be 0..4. 'Exactly 2 blue' means 3 red and 2 blue. 'At least 3 red' means 3 or 4 red (4 is the maximum). P(exactly r red, 5-r blue in first five) = C(4,r)C(6,5-r) / C(10,5), with C(10,5) = 252. P(3 red, 2 blue) = C(4,3)C(6,2)/252 = 4*15/252 = 60/252. P(4 red, 1 blue) = C(4,4)C(6,1)/252 = 1*6/252 = 6/252. Required conditional probability = P(3 red) / [P(3 red) + P(4 red)] = 60 / (60 + 6) = 60/66 = 10/11.

Example 2 (5 marks)

Bella throws a biased six-faced die whose faces are labelled 1, 2, 3, 4, 5 and 6. Every odd-numbered face carries the same probability, and every even-numbered face carries the same probability, with the chance of an even number being twice the chance of an odd number. (a) Determine the probability of rolling a 1. [2] Bella now throws the die three times. (b) Determine the probability that she obtains two 1s and one 2. [3]

Show the worked answer

(a) Let odd faces each have probability p and even faces each 2p. Total: 3p+3(2p)=1 => 9p=1 => p=1/9. So P(1)=1/9. (b) P(1)=1/9, P(2)=2/9. Two 1s and one 2 in three throws: the single '2' can be in any of 3 positions, so P = 3 * (1/9)² * (2/9) = 3 * (1/81)*(2/9) = 6/729 = 2/243.

Example 3 (2 marks)

Kitty's 10 uniquely designed charms (5 small, 3 medium, 2 large) are set in a line along a photo frame's base with no two small charms adjacent. In how many ways?

Show the worked answer

There are 5 small charms and 5 non-small charms (3 medium + 2 large), all distinct. First arrange the 5 non-small charms in a row: 5! = 120 ways. This creates 6 gaps (including the two ends) into which small charms can go so that no two smalls are adjacent. Place the 5 distinct small charms into 5 of these 6 gaps, one per gap: P(6,5) = 6*5*4*3*2 = 720 ways. Total = 120 * 720 = 86400.

More worked questions on this topic

Ask about ProbabilityUse Rae in Telegram

More A-Level H2 Maths topics

Functions · Graphs and their transformations · Equations and inequalities · Sequences and series · Vectors in two and three dimensions: basic properties · Scalar and vector products · all of A-Level H2 Maths