Probability
Conditional Probability
Grade 12
Question:
<p>A box contains 3 white and 2 red balls. If we draw one ball and without replacing the first ball, the probability of drawing red ball in the second draw is</p>
<p>(a) \(\frac{8}{25}\)</p>
<p>(b) \(\frac{2}{5}\)</p>
<p>(c) \(\frac{3}{5}\)</p>
<p>(d) \(\frac{21}{25}\)</p>
Step-by-Step Solution
Key Concept: Apply the law of total probability by conditioning on the color of the first ball drawn.
<p>Total balls = 5 (3 white, 2 red)</p><p>Case 1: First ball is white (prob \(\frac{3}{5}\)), then second is red (prob \(\frac{2}{4}\)): \(\frac{3}{5} \times \frac{2}{4} = \frac{6}{20}\)</p><p>Case 2: First ball is red (prob \(\frac{2}{5}\)), then second is red (prob \(\frac{1}{4}\)): \(\frac{2}{5} \times \frac{1}{4} = \frac{2}{20}\)</p><p>Total probability = \(\frac{6}{20} + \frac{2}{20} = \frac{8}{20} = \frac{2}{5}\)</p>
Correct Answer: B