<p>A dice is thrown three times and the sum of the thrown numbers is 15. Find the probability for which number 4 appears in first throw.</p>
Step-by-Step Solution
Key Concept: Use conditional probability P(A|B) = P(A∩B)/P(B), where A is 'sum equals 15' and B is '4 appears first'. Count favorable outcomes where first die shows 4 AND sum is 15, then divide by total outcomes summing to 15.
<p><strong>Step 1:</strong> Find all outcomes (a,b,c) where a+b+c=15 with 1≤a,b,c≤6.</p><p><strong>Step 2:</strong> List systematically: (3,6,6), (4,5,6), (4,6,5), (5,4,6), (5,5,5), (5,6,4), (6,3,6), (6,4,5), (6,5,4), (6,6,3)</p><p>Total outcomes with sum 15: <strong>10 outcomes</strong></p><p><strong>Step 3:</strong> Count outcomes where first die = 4: (4,5,6), (4,6,5)</p><p>Favorable outcomes: <strong>2 outcomes</strong></p><p><strong>Step 4:</strong> Apply conditional probability: P(4 appears first | sum is 15) = 2/10 = <strong>1/5</strong></p>
Correct Answer: Find P(A/B) where A: sum is 15, B: 4 appears first