NCEA Level 2 Mathematics 91267 Probability free practice questions with answers

Question 1 (13 marks) [Trees and conditional probability]

(a) In a town of 10000 people, 1% have a certain condition. A screening test correctly gives a positive result for 95% of people who have the condition, and correctly gives a negative result for 95% of people who do not. (i) How many people in the town would be expected to test positive? (ii) A person tests positive. Find the probability that they actually have the condition. (iii) Explain why this probability is lower than you might expect.

(a) Answer: (i) 590; (ii) 0.161; (iii) healthy people greatly outnumber those with the condition, so false positives swamp true positives

• Evidence level: Excellence (use a tree or natural frequencies, then interpret a conditional probability).

• People with the condition: 1% of 10000 = 100. Without: 9900.

• (i) True positives: 95% of 100 = 95. False positives: 5% of 9900 = 495. Expected positives = 95 + 495 = 590.

• (ii) P(condition | positive) = 95 ÷ 590 = 0.161.

• (iii) Most people do NOT have the condition, so even a small false-positive rate produces many false positives (495) compared with true positives (95).

(b) A bag contains 4 green marbles, 7 blue marbles and 6 yellow marbles. Two marbles are taken from the bag at random, one after the other, WITHOUT replacement. Find the probability that both marbles are the same colour.

(b) Answer: 0.309

• Evidence level: Merit (use a tree or list of cases without replacement).

• P(both the same) = P(both green) + P(both blue) + P(both yellow).

• = (

• = 0.309.

(c) A survey of 132 students produced the table below. (i) Find the probability that a randomly chosen person walk to school. (ii) Find the probability that a person are in a sports team, given that they walk to school. (iii) Use your answers to decide whether the two events are independent, and explain your decision.

(c) Answer: (i) 0.553; (ii) 0.671; (iii) not independent

• Evidence level: Excellence (compare a conditional probability with the unconditional one to test independence).

• (i) P(walk to school) = 4 17 )( 3 16 ) + ( 7 17 )( 6 16 ) + ( 6 17 )( 5 16 ). 73 132 = 0.553.

• (ii) Of the 73 who walk to school, 49 are in a sports team: P =

• (iii) Overall P(are in a sports team) =

• Here 0.671 ≠ 0.727, so the events are NOT independent (the probability changes when we know about the other event). 49 73 = 0.671. 96 132 = 0.727. Independent events have P(A | B) = P(A).