Box A contains 6 green balls and 3 yellow balls. Box B contains 4 green balls and x yellow balls. A ball is chosen at random from box A and placed in box B. A ball is then chosen at random from box B.
- (a)Draw a tree diagram to represent this information, showing the probability on each of the branches.[4]
- (b)The probability that both the balls chosen are the same colour is 8/15. Find the value of x.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [B1] Correct structure and probabilities for Box A branches.
- [B1] Completely correct structure and one correct probability for a Box B branch including label for G or Y.
- [B1] Completely correct structure and second correct probability on a Box B branch including label G or Y.
- [B1] Completely correct structure and remaining two probabilities correct on Box B branches, including labels for G or Y. SC B1 if correct shape diagram but only four correct algebraic probs for GG, GY, YG and YY.
- [M1] P(GG) + P(YY) = (6/9)*(5/(5+x)) + (3/9)*( (x+1)/(5+x) ) (or) x their (5+x+or their (1+x))
- [M1] 15(x+11) = 24(x+5) OE Accept sum of their products equated to 8/15 and rearranged to form a linear equation.
- [A1] Solve: x = 5
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