One card is missing from a pack of cards. Let $A$ be the event that missing card is a spade. Then two cards are drawn, and $S$ be the event that they are spades then:
Step-by-Step Solution
Key Concept: Apply Bayes' theorem with conditional probability: P(A|S) = P(S|A)·P(A)/P(S), where P(S|A) = C(12,2)/C(50,2) when one spade is missing, and P(S|A') = C(13,2)/C(50,2) when a non-spade is missing. The total probability P(S) = P(S|A)·P(A) + P(S|A')·P(A').
For event $A'$ (missing card is not a spade), there are 39 non-spade cards out of 52, so $P(A') = \frac{39}{52} = \frac{3}{4}$. For event $S$ (missing card is spade), only 12 spades remain out of 51 cards, giving $P(S|A) = \frac{12}{51}$. Using conditional probability, $P(A|S) = \frac{P(A)P(S|A)}{P(S)} = \frac{\frac{1}{4} \cdot \frac{12}{51}}{\frac{13}{51}} = \frac{11}{50}$.
Correct Answer: 1,3