Probability
Conditional Probability
Grade 12
Question:
<p>Two cards are drawn without replacement from a well-shuffled pack. The probability that one of them is an ace of heart, is</p>
<p>(a) \(\frac{1}{25}\)</p>
<p>(b) \(\frac{1}{26}\)</p>
<p>(c) \(\frac{1}{52}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by considering the two mutually exclusive cases for when the ace of heart appears.
<p>The probability that one of the two cards is the ace of heart.</p><p>Case 1: First card is ace of heart, second is not: \(P_1 = \frac{1}{52} \times \frac{51}{51} = \frac{1}{52}\)</p><p>Case 2: First card is not ace of heart, second is ace of heart: \(P_2 = \frac{51}{52} \times \frac{1}{51} = \frac{1}{52}\)</p><p>Total probability = \(P_1 + P_2 = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}\)</p>
Correct Answer: B