Unit 10
9 marks · NormalThis question connects algebraic surds with conditional probability.
Leo is working with the expression . Decide whether is rational or irrational. Explain your answer.
A numbered spinner has 5 sections: 2 labelled 'A' and 3 labelled 'B'. Each time it is spun, the outcome is equally likely.
If the spinner is spun twice, find, with reasons, the probability of getting 'A' on both spins. Write your answer as a fraction in its lowest terms.
Expand the expression and comment on whether the result is rational or irrational.
The spinner is spun twice. Given that 'A' appeared at least once, what is the probability that the first spin was 'A'? Give your reasoning and write your answer as a fraction in its simplest form.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) x is irrational because it is a sum of a rational and an irrational number.
(b) (2/5) × (2/5) = 4/25
(c) (1+√6)(1−√6)=1−(√6)^2=1−6=−5; result is rational.
(d) Let E = 'A appears at least once'; possible outcomes: AA, AB, BA. Probability 1st A and at least one A: (AA, AB). P(AA)=4/25, P(AB)=6/25. P(E)=13/25. Required probability = (P(AA)+P(AB))/P(E) = (4/25+6/25)/(13/25) = 10/13.
Common mistakes: Confusing rational/irrational results of surd expressions; Wrong denominator for conditional probability
Show mark scheme and worked solution
Marking points
B1 (a) · 2 marks · Correct explanation of irrationality for surd expression.
M1 (b) · 1 mark · Multiplies probabilities correctly for two independent events. (follow-through)
A1 (b) · 1 mark · Gives answer as 4/25. (follow-through)
B1 (c) · 2 marks · Expands expression correctly and identifies result as rational.
M1 (d) · 1 mark · Correct set-up of conditional probability using at least one A. (follow-through)
A1 (d) · 1 mark · Numerator and denominator in conditional calculation are correct. (follow-through)
A1 (d) · 1 mark · Final probability as 10/13 (simplest form). (follow-through)
Worked solution
(a) x = 5−2√6. 2√6 is irrational and 5 is rational; sum is irrational.
(b) Probability of A on one spin = 2/5. Both spins: (2/5)*(2/5)=4/25.
(c) (1+√6)(1−√6) = 1−(√6)^2 = 1−6 = −5. Result is rational.
(d) The possible outcomes are AA, AB, BA, BB. P(AA)=4/25; P(AB)=6/25; P(BA)=6/25; P(BB)=9/25.
The event 'at least one A' excludes BB, so total probability is 1−9/25=16/25. (But listing shows AA, AB, BA: 3 outcomes, but with probabilities as above.)
Probability first spin is A and at least one A: AA and AB: 4/25 + 6/25 = 10/25. Total P(at least one A) = 16/25.
So required probability = 10/25 ÷ 16/25 = 10/16 = 5/8.