A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

Solution

Maths sir · 3 steps

Split on what the missing card was, then let the two spades update the odds through Bayes.

1. Two cases. Let E₁ be the case where the missing card is a spade and E₂ the case where it is not. Before any drawing, P(E₁) = 13/52 = 1/4 and P(E₂) = 3/4.
2. Two spades drawn. There are 51 cards left. If a spade is missing, 12 spades remain, so P(A | E₁) = ¹²C₂/⁵¹C₂ = 66/1275. If a non-spade is missing, all 13 spades remain, so P(A | E₂) = ¹³C₂/⁵¹C₂ = 78/1275.

   P(E₂ | A) = (3/4 × 78/1275)/(1/4 × 66/1275 + 3/4 × 78/1275) = 234/(66 + 234) = 234/300

3. So. The common factors 1/4 and 1/1275 cancel, leaving 234/300 = 39/50. The answer is (a).

The idea to keep. In Bayes problems, first list the hidden cases with their prior probabilities. Then find how likely the observed event is in each case. Factors common to every term cancel, so do not expand them.

Answer from the NTA final answer key. Solution by PrepWiser.