<p>One ticket is selected at random from 50 tickets numbered 00, 01, 02, … 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(A|B) = P(A∩B)/P(B). First identify all tickets where product of digits = 0 (contains at least one 0), then count how many of these also have digit sum = 8.
<p><strong>Step 1:</strong> Identify the sample space for 'product of digits = 0'. The ticket must contain at least one 0.</p><p>Tickets with 0 in units place: 00, 10, 20, 30, 40 (5 tickets)</p><p>Tickets with 0 in tens place: 01, 02, 03, 04, 05, 06, 07, 08, 09 (9 tickets)</p><p>Total tickets with product = 0: {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40} = 14 tickets</p><p><strong>Step 2:</strong> Among these 14 tickets, find those with digit sum = 8.</p><p>From units-place 0 group: 00(0), 10(1), 20(2), 30(3), 40(4) — none sum to 8</p><p>From tens-place 0 group: 01(1), 02(2), 03(3), 04(4), 05(5), 06(6), 07(7), 08(8)✓, 09(9)</p><p>Only ticket 08 has digit sum = 8.</p><p><strong>Step 3:</strong> Apply conditional probability formula.</p><p>P(sum = 8 | product = 0) = (Number of tickets with both sum=8 AND product=0)/(Total tickets with product=0)</p><p>= 1/14</p><p>∴ Answer: A</p>
Correct Answer: A