Probability
Conditional Probability and Total Probability Law
Grade 12
Question:
<p>A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is</p>
<p>(a) \(\frac{3}{10}\)</p>
<p>(b) \(\frac{2}{5}\)</p>
<p>(c) \(\frac{1}{5}\)</p>
<p>(d) \(\frac{3}{4}\)</p>
Step-by-Step Solution
Key Concept: Use the Law of Total Probability by conditioning on the colour of the first ball drawn, accounting for how the composition changes after replacement.
<p><strong>Solution:</strong> Let $E_1$ = Event that first ball drawn is red, and $E_2$ = Event that first ball drawn is black.</p><p>Initial: 4 red and 6 black balls (total 10).</p><p>$P(E_1) = \frac{4}{10} = \frac{2}{5}$</p><p>$P(E_2) = \frac{6}{10} = \frac{3}{5}$</p><p><strong>Case 1:</strong> If first ball is red, it is returned with 2 more red balls. Now bag has 6 red and 6 black (total 12).</p><p>$P(\text{Second ball red} | E_1) = \frac{6}{12} = \frac{1}{2}$</p><p><strong>Case 2:</strong> If first ball is black, it is returned with 2 more black balls. Now bag has 4 red and 8 black (total 12).</p><p>$P(\text{Second ball red} | E_2) = \frac{4}{12} = \frac{1}{3}$</p><p><strong>By Law of Total Probability:</strong></p><p>$P(\text{Second ball red}) = P(E_1) \cdot P(\text{Red}|E_1) + P(E_2) \cdot P(\text{Red}|E_2)$</p><p>$= \frac{2}{5} \cdot \frac{1}{2} + \frac{3}{5} \cdot \frac{1}{3}$</p><p>$= \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$</p><p>∴ Answer is (b) $\frac{2}{5}$</p>
Correct Answer: b