Probability
Conditional Probability
Grade None
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>\(\dfrac{2}{5}\)</p>
<p>\(\dfrac{1}{5}\)</p>
<p>\(\dfrac{3}{4}\)</p>
<p>\(\dfrac{3}{10}\)</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by conditioning on the first draw. After the first draw and replacement protocol, the bag composition changes, affecting the second draw's probability.
<p><strong>Step 1: Identify the initial state</strong></p><p>Initially: 4 red balls, 6 black balls (total 10 balls)</p><p><strong>Step 2: Apply law of total probability by cases</strong></p><p>Let R₁ = first ball drawn is red, B₁ = first ball drawn is black</p><p>P(R₁) = 4/10 = 2/5, P(B₁) = 6/10 = 3/5</p><p><strong>Step 3: Determine bag composition after first draw</strong></p><p><strong>Case 1:</strong> If first ball is RED (prob = 2/5):<br/>→ Return 1 red + 2 additional red balls<br/>→ New composition: 7 red, 6 black (total 13 balls)<br/>→ P(2nd ball is red | R₁) = 7/13</p><p><strong>Case 2:</strong> If first ball is BLACK (prob = 3/5):<br/>→ Return 1 black + 2 additional black balls<br/>→ New composition: 4 red, 8 black (total 12 balls)<br/>→ P(2nd ball is red | B₁) = 4/12 = 1/3</p><p><strong>Step 4: Calculate total probability</strong></p><p>P(2nd ball is red) = P(R₁) × P(2nd red | R₁) + P(B₁) × P(2nd red | B₁)</p><p>= (2/5) × (7/13) + (3/5) × (1/3)</p><p>= 14/65 + 3/15</p><p>= 14/65 + 13/65</p><p>= 27/65</p><p><strong>∴ Answer: A (27/65)</strong></p>
Correct Answer: A