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Cambridge International AS & A Level · Topic 5.3
A-Level Maths 9709 / Topic 5.3 / AS & A Level

Count it, add it, multiply it.

Paper 5 probability rests on a few moves: count equally likely outcomes, add for 'or' when events are exclusive, multiply along the branches of a tree for 'and', and use combinations when several items are picked at once.

Full-screen labCombined probability lab → Topic 5.3All probability lessons →
The key idea

For equally likely outcomes, \(\text{P}(A) = \dfrac{\text{number of outcomes in } A}{\text{total number of outcomes}}\). If \(A\) and \(B\) are exclusive, \(\text{P}(A \cup B) = \text{P}(A) + \text{P}(B)\). Along a tree, multiply; across separate routes, add. \(\text{P}(\text{at least one}) = 1 - \text{P}(\text{none})\).

RB RBRB 5/83/8 4/73/75/72/7 RR: 20/56RB: 15/56BR: 15/56BB: 6/56 1st ball2nd ball Same colour: 20/56 + 6/56 = 26/56 = 13/28
Fig. 1: two balls taken without replacement from a bag of 5 red and 3 blue. The second-stage probabilities change because one ball has gone; the four end results add up to 1
Section 01

Equally likely outcomes.

In the lab's sample-space tab, roll two dice many times and compare the frequency of each total with the grid of 36 outcomes. When every outcome in the sample space is equally likely, probability is just counting.

Worked example Totals of two dice
Two fair dice are thrown. Find the probability that (a) the total is 8, (b) the total is at least 10.
  1. The sample space has \(6 \times 6 = 36\) equally likely outcomes.
  2. (a) Total 8: (2,6), (3,5), (4,4), (5,3), (6,2), which is 5 outcomes.
  3. (b) Total 10: 3 outcomes; 11: 2; 12: 1. That is 6 outcomes.
Answer: (a) \(\dfrac{5}{36}\)   (b) \(\dfrac{6}{36} = \dfrac{1}{6}\)
Your turn
Two fair dice are thrown. Find the probability that the product of the scores is odd.
Examiner trap

Saying P(total 8) \(= \dfrac{1}{11}\) because there are 11 possible totals. The totals 2 to 12 are NOT equally likely; only the 36 ordered pairs are.

Section 02

Exclusive and independent.

Two words that are easily confused. Exclusive events cannot happen together, so for 'A or B' you add. Independent events do not affect each other, so for 'A and B' you multiply.

ExclusiveIndependent
\(\text{P}(A \cap B) = 0\)\(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)\)
\(\text{P}(A \cup B) = \text{P}(A) + \text{P}(B)\)One happening does not change the probability of the other
e.g. one die shows 2, the same die shows 5e.g. one die shows 2, a second die shows 5
Worked example Using both rules
\(\text{P}(A) = 0.3\) and \(\text{P}(B) = 0.5\). Find \(\text{P}(A \cup B)\) if (a) \(A\) and \(B\) are exclusive, (b) \(A\) and \(B\) are independent.
  1. (a) Exclusive: \(0.3 + 0.5 = 0.8\).
  2. (b) Independent: \(\text{P}(A \cap B) = 0.3 \times 0.5 = 0.15\).
  3. Neither happens: \(0.7 \times 0.5 = 0.35\), so \(\text{P}(A \cup B) = 1 - 0.35 = 0.65\).
Answer: (a) 0.8   (b) 0.65
Worked example At least one
A fair die is thrown 4 times. Find the probability of at least one six.
  1. Listing every case with one, two, three or four sixes is long. Use the complement.
  2. No sixes: \(\left(\dfrac{5}{6}\right)^4 = \dfrac{625}{1296}\), since the throws are independent.
  3. \(1 - \dfrac{625}{1296} = \dfrac{671}{1296}\).
Answer: \(\dfrac{671}{1296} = 0.518\) (3 s.f.)
Examiner trap

Adding probabilities of events that are NOT exclusive. With \(\text{P}(A) = 0.6\) and \(\text{P}(B) = 0.7\), 'adding' gives 1.3, which is impossible; the overlap has been counted twice.

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Two fair dice are thrown. What is the probability that the total is 8?

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\(A\) and \(B\) are exclusive with \(\text{P}(A) = 0.3\) and \(\text{P}(B) = 0.45\). Find \(\text{P}(A \cup B)\).

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\(A\) and \(B\) are independent with \(\text{P}(A) = 0.3\) and \(\text{P}(B) = 0.45\). Find \(\text{P}(A \cap B)\).

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Three fair coins are thrown. What is the probability of at least one head?

Section 03

Tree diagrams.

Use the lab's tree-diagram tab to build a 'without replacement' tree, then compare it with Fig. 1. Each branch carries the probability given what has already happened; multiply along a route and add the routes you want.

Worked example Without replacement
A bag contains 5 red and 3 blue balls. Two are taken at random without replacement. Find the probability that (a) they are the same colour, (b) at least one is blue.
  1. (a) RR: \(\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56}\). BB: \(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56}\).
  2. Same colour: \(\dfrac{20}{56} + \dfrac{6}{56} = \dfrac{26}{56}\).
  3. (b) At least one blue is everything except RR: \(1 - \dfrac{20}{56} = \dfrac{36}{56}\).
Answer: (a) \(\dfrac{13}{28}\)   (b) \(\dfrac{9}{14}\)
Worked example A stopping process
Raj has up to 3 attempts at a puzzle and stops as soon as he succeeds. The probability that he succeeds is 0.4 on the first attempt, 0.5 on the second and 0.6 on the third. Find the probability that he succeeds.
  1. Succeeds 1st time: 0.4.
  2. Fails, then succeeds: \(0.6 \times 0.5 = 0.3\).
  3. Fails twice, then succeeds: \(0.6 \times 0.5 \times 0.6 = 0.18\).
  4. Add the routes: \(0.4 + 0.3 + 0.18\).
Answer: 0.88 (check: fails all three, \(0.6 \times 0.5 \times 0.4 = 0.12\), and \(1 - 0.12 = 0.88\))
Examiner trap

Keeping the same denominators on the second branches. Without replacement there are only 7 balls left, so the second-stage fractions are sevenths.

Section 04

Probability by combinations.

When several items are picked at once, count the favourable selections and the total selections with \({}^{n}C_{r}\). This avoids drawing a large tree and automatically covers every order.

Worked example Exactly two red
Three balls are taken at random from a bag of 5 red and 3 blue. Find the probability that exactly two are red.
  1. Total selections: \({}^{8}C_{3} = 56\).
  2. Favourable: 2 red from 5 AND 1 blue from 3: \({}^{5}C_{2} \times {}^{3}C_{1} = 10 \times 3 = 30\).
  3. \(\dfrac{30}{56}\). Check with a tree: \(3 \times \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} = \dfrac{180}{336} = \dfrac{15}{28}\).
Answer: \(\dfrac{15}{28}\)
Worked example Arrangements at random
The letters of NEEDLESS are arranged in a random order. Find the probability that the arrangement begins and ends with S.
  1. Total arrangements: \(\dfrac{8!}{3!\,2!} = 3360\). Beginning and ending with S: \(\dfrac{6!}{3!} = 120\).
  2. \(\dfrac{120}{3360} = \dfrac{1}{28}\). Check directly: \(\dfrac{2}{8} \times \dfrac{1}{7} = \dfrac{1}{28}\).
Answer: \(\dfrac{1}{28}\)
Examiner trap

Forgetting the different orders when using a tree for 'exactly two red'. RRB alone gives \(\dfrac{5}{28}\); there are 3 orders (RRB, RBR, BRR). Combinations count them automatically.

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A bag contains 5 red and 3 blue balls. Two are taken at random without replacement. Find the probability that they are the same colour.

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A fair die is thrown 4 times. Find the probability of at least one six.

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Three balls are taken at random from a bag of 6 red and 4 blue balls. Find the probability that exactly two are red.

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Two fair dice are thrown. Find the probability that the total is a prime number.

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Six original Cambridge-style questions
Sample spaces for two dice, exclusive and independent events, without-replacement trees, at least one, combinations and a stopping process.