For her bedtime drink, Suki has either chocolate, tea or milk with probabilities 0.45, 0.35 and 0.2 respectively. When she has chocolate, the probability that she has a biscuit is 0.3. When she has tea, the probability that she has a biscuit is 0.6. When she has milk, she never has a biscuit. Find the probability that Suki has tea given that she does not have a biscuit.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

1. [M1] 0.45×a+0.35×b+0.2[×1],a=0.7,0.3b = 0.4,0.6, seen anywhere.
2. [A1] Correct, accept unsimplified.
3. [M1] P(T∩B') = 0.35×0.4 [= 0.14, 7/50] Seen as numerator or denominator of a fraction.
4. [M1] Values substituted into conditional probability formula correctly. Accept unsimplified. Denominator sum of 3 two-factor probabilities (condone omission of 1 from final factor). If clearly identified, condone from incomplete denominator.
5. [A1] 0.214, 28/131 If 0 marks awarded, SC B1 0.214 WWW.