Probability
Classical Probability
Grade 12
Question:
<p>In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of 3 girls sitting together in a back row on adjacent seats?</p>
<p>\(\dfrac{1}{91}\)</p>
<p>\(\dfrac{2}{91}\)</p>
<p>\(\dfrac{1}{182}\)</p>
<p>\(\dfrac{3}{91}\)</p>
Step-by-Step Solution
Key Concept: Recognize that 'three girls sitting together in a back row on adjacent seats' is a constrained arrangement problem where we treat the 3 girls as a unit in specific positions, then use conditional probability: P = (Favorable arrangements)/(Total arrangements).
<p><strong>Step 1: Total seating arrangements</strong></p><p>We have 12 people and 14 seats total (2 vans × 7 seats). We need to choose 12 seats from 14 and arrange 12 people: <strong>P(14,12) = 14!/(14-12)! = 14!/2!</strong></p><p><strong>Step 2: Favorable arrangements (girls sitting together in back row on adjacent seats)</strong></p><p>Each van has 4 back seats. Adjacent triples in a back row: positions (1,2,3) or (2,3,4) = <strong>2 positions per van × 2 vans = 4 possible blocks</strong></p><p>For each block:</p><ul><li>Arrange 3 girls within the block: 3! ways</li><li>Arrange remaining 9 boys in remaining 11 seats: P(11,9) = 11!/2! ways</li></ul><p><strong>Favorable = 4 × 3! × P(11,9) = 4 × 6 × (11!/2!)</strong></p><p><strong>Step 3: Calculate probability</strong></p><p>P = (4 × 3! × 11!/2!) / (14!/2!)</p><p>P = (4 × 6 × 11!) / 14!</p><p>P = (24 × 11!) / (14 × 13 × 12 × 11!)</p><p>P = 24 / (14 × 13 × 12)</p><p>P = 24 / 2184 = <strong>1/91</strong></p><p>∴ Answer: A</p>
Correct Answer: A