Past Paper exam Question which i need help on.
Two colours of sweets; Red and Green. The two types are mixed into bags by one of two mixing machines. Mixer A produces 40% red, and 60% green. Mixer B produces 70% red and 30% green. Mixer A is used 30% of the time.
(i) What is the probability the 1st sweet chosen is red?
(ii) If it is red, what is the probability the packet was mixed by machine A?
(iii) If the second sweet from the same packet is also red, what now is the probability the packet was mixed by machine A?
(i) So for this, ive got;
P(R) = P(R|A)P(A) + P(R|B)P(B)
P(R) = 0.4 x 0.3 + 0.7 x 0.7 = 0.61
P(A|R) = [P(R|A)P(A)]/P(R)
P(A|R) = 0.12/0.61 = 12/61
Is this correct so far? How do i go about working out part (iii)?