John removes the labels from three cans of tomato soup and two cans of chicken soup in order to enter a competition, and puts the cans away. He then discovers that the cans are identical, so that he cannot distinguish between cans of tomato soup and chicken soup. Some weeks later he decides to have a can of chicken soup for lunch. He opens the cans at random until he opens a can of chicken soup. Let Y denote the number of cans he opens.

Find

(a)     the possible values of Y ,

(b)     the probability of each of these values of Y ,

(c)     the expected value of Y .

(a)     1, 2, 3, 4     A1

(b)     \({\text{P}}(Y = 1) = \frac{2}{5}\)     A1

\({\text{P}}(Y = 2) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{{10}}\)     A1

\({\text{P}}(Y = 3) = \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} = \frac{1}{5}\)     A1

\({\text{P}}(Y = 4) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} = \frac{1}{{10}}\)     A1

(c)     \({\text{E}}(Y) = 1 \times \frac{2}{5} + 2 \times \frac{3}{{10}} + 3 \times \frac{1}{5} + 4 \times \frac{1}{{10}}\)     M1

\( = 2\)     A1

[7 marks]

Candidates found this question challenging with only better candidates gaining the correct answers. A number of students assumed incorrectly that the distribution was either Binomial or Geometric.