WJEC Maths for A2 – Applied

1 Probability 18 Converting the events to A and B notation we obtain P( A′ ∩ B ) = P( A′  ) × P( B  | A′  ) = 0.40 × 0.97 = 0.388 (b) Pȍ B Ȏ = Pȍ A Ȏ × Pȍ B | A Ȏ + Pȍ A’  Ȏ × Pȍ B | A’ Ȏ = 0.60 × 0.32 + 0.40 × 0.97 = 0.58 (c) Pȍ A ∩ B Ȏ = Pȍ A Ȏ × Pȍ B | A Ȏ = 0.60 × 0.32 = 0.192 Now P( A | B ) = Pȍ A ∩ B Ȏ Pȍ B Ȏ = 0.192 0.58 = 0.33 (2 s.f.) 1.3 Use of Venn diagrams for conditional probability One way of solving probability questions is to ϐirst draw a Venn diagram. The drawing of Venn diagrams was looked at in Topic 3 of the AS Applied book. You should look back at this topic before continuing with this new topic. Work through the following example to check your familiarity with Venn diagrams. Examples 1 The probabilities for two events A and B are shown in the following Venn diagram. B A S 0.55 0.2 0.1 0.15 Find: (a) P( A ) (b) P( A only) (c) P( A ∩ B ) (d) P( A ∪ B ) (e) P( A ′ ∩ B ) You found the probability that the student is telling the truth Pȍ B Ȏ in part (b), so you can use this in the formula. (i.e. Pȍ B Ȏ = 0.58) Find Pȍ A | B Ȏ by substituting your values for P( A ∩ B ) and Pȍ B Ȏ into the conditional probability formula.