A-Level Discrete random variables: worked solution

Question

A bag holds 6 red, 6 green, 6 yellow and 6 blue discs, identical except for colour. Four discs are removed successively without replacement. Let X be the count of distinct colours seen among those four discs. (a) Prove that P(X=2)=465/1771.

There are 6 + 6 + 6 + 6 = 24 discs, and 4 are taken without replacement. The order of drawing does not affect the set of colours seen, so count unordered selections: Total selections = C(24,4) = (24 × 23 × 22 × 21)/(4 × 3 × 2 × 1) = 10626 All of these are equally likely, since the discs are identical apart from colour. X = 2 means the 4 discs show EXACTLY two different colours. Step 1 - choose which two colours appear: C(4,2) = 6 ways Step 2 - for a fixed pair of colours there are 6 + 6 = 12 discs available, and we take 4 of them, but we must use BOTH colours (using only one would give X = 1, not 2). All selections of 4 from those 12: C(12,4) = (12 × 11 × 10 × 9)/(4 × 3 × 2 × 1) = 495 Subtract the selections that use only ONE of the two colours: all 4 from the first colour, C(6,4) = 15, or all 4 from the second colour, C(6,4) = 15. Selections using both colours = 495 − 15 − 15 = 465 Step 3 - combine. The pair of colours and then the selection within that pair are chosen in sequence, so multiply: Favourable selections = 6 × 465 = 2790 P(X = 2) = 2790/10626 Divide numerator and denominator by 6: 2790 ÷ 6 = 465 and 10626 ÷ 6 = 1771 P(X = 2) = 465/1771 (as required)