Conditional probability — Notes & Worked ExamplesShow / Hide
Explanation:
Conditional probability occurs when one event changes the probabilities for the next event, most commonly when items are chosen without replacement. On a tree diagram, always reduce both the total denominator by 1 on the second set of branches and the numerator by 1 for whichever item was picked first.
Multiply along branches for AND, and add the results of separate paths for OR. For "at least one" or "two of the same colour", you can also use 1 − P(opposite event) as a fast check. For Venn diagrams or two-way tables, restrict the denominator strictly to the "given" group.
Example 1: A bag has 5 red and 3 blue counters. Two are taken at random without replacement. Find P(one of each colour).
Work out In a group of students, 11 study French only, 9 study Spanish only, and 5 study both French and Spanish. A student who studies Spanish is chosen at random. Find the conditional probability that this student also studies French.
Worked Solution & Answer
P(F \mid S) = n(F ∩ S)n(S) = 55 + 9 = 514B1
Answer:514
Question 2
[1 mark]
Work out In a group of students, 10 study French only, 10 study Spanish only, and 9 study both French and Spanish. A student who studies Spanish is chosen at random. Find the conditional probability that this student also studies French.
Worked Solution & Answer
P(F \mid S) = n(F ∩ S)n(S) = 99 + 10 = 919B1
Answer:919
Question 3OCR Nov 2022 4H Q16
[2 marks]
A salesroom sells various types of car. Some cars are electric (E), some are petrol (P), some are both and some are neither. The Venn diagram below shows the salesroom’s stock of cars.
A petrol car is picked at random. Find the probability that the car is also electric.
Worked Solution & Answer
Total petrol cars = 17 + 8 = 25 M1 Number of petrol cars that are also electric = 17 Probability = 17 / 25 (or 0.68). A1
Vicky has a collection of medals. The Venn diagram gives information about the number of medals in her collection where [INDENT]\mathscr{E} = {all medals} [INDENT]A = {English medals} [INDENT]B = {gold medals}
Vicky is going to take at random a medal from her collection. Given that the medal is gold, the probability that the medal is English is 211 Work out the number of medals in Vicky’s collection.
A school year has 78 students. 28 wear glasses. 14 of the students who wear glasses are left-handed. 30% of the students who do **not** wear glasses are left-handed.
(a)ξ = students in the school year G = wears glasses L = left-handed Complete the Venn diagram.[3 marks]
(b)A left-handed student is chosen at random. Work out the probability that the student wears glasses.[1 mark]
Answer
Worked Solution & Answer
(a)
Both = 1/4 × 28 = 7. G only = 28 – 7 = 21 M1 Do not wear glasses = 78 – 28 = 50 M1 L only = 30% of 50 = 15 Outside = 50 – 15 = 35 A1
Answer (a):G only = 21, Intersection = 7, L only = 15, Outside = 35
The Venn diagram shows information about 80 people who visited an online shop. ξ = 80 people T = people who bought trainers H = people who bought a hoodie
(a)One person is chosen at random. Work out the probability that they bought a hoodie.[3 marks]
Answer
(b)One person who bought trainers is chosen at random. Work out the probability that they bought a hoodie.[1 mark]
Answer
Worked Solution & Answer
(a)
15 + 8 + x + 2x = 80 ⇒ 23 + 3x = 80 ⇒ 3x = 57 ⇒ x = 19 M2 Number who bought a hoodie = 8 + 19 = 27 ⇒ P(H) = 27/80 A1
Answer (a): Answer 2780
(b)
Number who bought trainers = 15 + 8 = 23 ⇒ P(H | T) = 8/23 B1
The Venn diagram shows information about some houses. G = houses with a garage S = houses with a shed A house is chosen at random.
(a)The house has a garage. What is the probability that it has a shed?[1 mark]
Answer
(b)The house does not have a garage. What is the probability that it does not have a shed?[1 mark]
Answer
(c)Show that P(G ∩ S)' > P(G ∪ S')[2 marks]
Worked Solution & Answer
(a)
Houses with a garage = 13 + 8 = 21 Of these, 8 have a shed Probability = 821. B1
Answer (a): Answer 821
(b)
Houses without a garage = 11 + 4 = 15 Of these, 4 do not have a shed Probability = 415. B1
Answer (b): Answer 415
(c)
Total houses = 13 + 8 + 11 + 4 = 36. M1 G ∩ S = 8 ⇒ P((G ∩ S)') = 36 - 836 = 2836 = 79. G ∪ S' includes houses with a garage (21) plus houses with neither (4): Total in G ∪ S' = 13 + 8 + 4 = 25 ⇒ P(G ∪ S') = 2536 Since 2836 > 2536, P(G ∩ S)' > P(G ∪ S'). C1
A school has 90 students in Year 11. • 50 of the students study French. • 59 of the students study Spanish. • 4 of the students do not study either of these languages. One of the students is chosen at random. Find the probability that the student studies French given that they study Spanish. You may use the grid **or** the Venn diagram to help present your working if you wish.
Worked Solution & Answer
Number studying French or Spanish = 90 − 4 = 86 M2 Number studying both French and Spanish = 50 + 59 − 86 = 23 M2 P(French | Spanish) = 23 / 59. A1
A shop manager wants to advertise special offers on social media platforms. The manager asks 100 customers which of type A, type B or type C they use. Of these customers, [INDENT]4 use all three types [INDENT]16 do not use any of type A, type B or type C [INDENT]8 use both type A and type B, but not type C [INDENT]14 use both type B and type C [INDENT]62 in total use type A [INDENT]all 20 who use type C also use at least one of type A and type B.
(a)Complete the Venn diagram for this information.[4 marks]
(b)find the probability that this customer also uses type B.[2 marks]
Worked Solution & Answer
(a)
B∩C only = 14 – 4 = 10 M1 C only = 0 M1 A∩C only = 20 – 14 = 6 M1 A only = 62 – (8 + 4 + 6) = 44 B only = 100 – 16 – 62 – 10 = 12 A1
Answer (a):A only = 44, B only = 12, C only = 0, A∩B only = 8, B∩C only = 10, A∩C only = 6, A∩B∩C = 4, outside = 16
