O-LevelMathematicsProbability4024/21 · Oct/Nov 2020 · Q810 Marks

A bag contains 36 balls. There are n red balls in the bag. The rest of the balls are green. Esther takes two balls from the bag, at random, without replacement.

(a)Complete the tree diagram.[2]
(b)Write an expression, in terms of n, for the probability that Esther's first ball is red and her second ball is green.[1]
(c)The probability that Esther's first ball is red and her second ball is green is 1/7. Show that n² - 36n + 180 = 0.[2]
(d)Solve the equation n²-36n+180 = 0. Show your working.[2]
(e)There are more green balls than red balls in the bag. Find the probability that Esther takes two green balls. Give your answer as a fraction in its lowest terms.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

[2] n/35 and 35-n/35 oe correctly positioned on tree diagram B1 for each
[1] n/36 x (36-n)/35 oe
[2] Correct rearrangement to n² - 36n + 180 = 0 AG M1 FT their (b) provided both probabilities in terms of n A1 A0 if any errors or omissions in working
[2] (n-6)(n-30) [= 0] B1 Or 36 +- √36^2 - 4*180 / 2 6,30 B1
[3] 29/42 final answer B1 for use of n = 6 M1 for (36-their 6)/36 x (35-their 6)/35 oe

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