Probability
Conditional Probability
Grade 12
Question:
<p>An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is __________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by conditioning on the color of the first ball drawn. After each draw, the composition changes before the second draw, requiring careful tracking of urn states.
<p><strong>Step 1: Set up initial state</strong></p><p>Initial urn: 5 red, 2 green (total 7 balls)</p><p><strong>Step 2: Case 1 - First ball is GREEN</strong></p><p>P(1st green) = 2/7</p><p>After drawing green (not returned) and adding red:</p><p>Urn becomes: 5+1=6 red, 2-1=1 green (total 7 balls)</p><p>P(2nd red | 1st green) = 6/7</p><p><strong>Step 3: Case 2 - First ball is RED</strong></p><p>P(1st red) = 5/7</p><p>After drawing red (not returned) and adding green:</p><p>Urn becomes: 5-1=4 red, 2+1=3 green (total 7 balls)</p><p>P(2nd red | 1st red) = 4/7</p><p><strong>Step 4: Apply Law of Total Probability</strong></p><p>P(2nd red) = P(2nd red | 1st green)·P(1st green) + P(2nd red | 1st red)·P(1st red)</p><p>P(2nd red) = (6/7)·(2/7) + (4/7)·(5/7)</p><p>P(2nd red) = 12/49 + 20/49 = 32/49</p><p><strong>Step 5: Convert to decimal</strong></p><p>32/49 = 0.653061...</p><p>∴ Answer: <strong>0.6531</strong></p><p><em>Note: If answer given is 0.5238, verify problem statement. For standard interpretation above, answer is 0.6531</em>
Correct Answer: 0.5238