Probability
Permutation-based Probability
Grade 12
Question:
<p>There are two vans each having numbered seats, 3 in the front and 4 at the back. There are 3 girls and 9 boys to be seated in the vans. The probability of 3 girls sitting together in a back row on adjacent seats, is</p>
<p>(a) \(\frac{1}{13}\)</p>
<p>(b) \(\frac{1}{39}\)</p>
<p>(c) \(\frac{1}{65}\)</p>
<p>(d) \(\frac{1}{91}\)</p>
Step-by-Step Solution
Key Concept: We need to find the probability that all 3 girls sit together in adjacent seats in a back row. This requires counting favorable arrangements (3 girls in one back row, adjacent seats) and dividing by total arrangements of 12 people in 14 seats.
<p><strong>Step 1: Count total seats and arrangement.</strong></p><p>Total seats = 2 vans × (3 front + 4 back) = 14 seats</p><p>We need to arrange 12 people (3 girls + 9 boys) in 14 seats.</p><p>Total arrangements = P(14,12) = 14!/(14-12)! = 14!/2! = 14 × 13 × 12!</p><p></p><p><strong>Step 2: Identify back rows where 3 girls can sit together.</strong></p><p>Each van has one back row with 4 numbered seats. Total of 2 back rows available.</p><p>In a row of 4 seats, there are exactly 2 ways to choose 3 adjacent seats: {1,2,3} or {2,3,4}</p><p>Total positions for 3 adjacent girls in back rows = 2 vans × 2 positions per row = 4 positions</p><p></p><p><strong>Step 3: Count favorable arrangements.</strong></p><p>For each of the 4 positions where 3 girls sit adjacent:</p><p>- Arrange 3 girls in these 3 seats: 3! = 6 ways</p><p>- The remaining 9 boys and 1 empty seat must be arranged in remaining 11 seats: P(11,9) = 11!/2! = 11 × 10 × 9!</p><p></p><p>Favorable arrangements = 4 × 3! × P(11,9) = 4 × 6 × (11 × 10 × 9!)</p><p></p><p><strong>Step 4: Calculate probability.</strong></p><p>P = (4 × 6 × 11 × 10 × 9!) / (14 × 13 × 12!)</p><p></p><p>P = (4 × 6 × 11 × 10 × 9!) / (14 × 13 × 12 × 11 × 10 × 9!)</p><p></p><p>P = (4 × 6) / (14 × 13 × 12)</p><p></p><p>P = 24 / (14 × 13 × 12) = 24 / 2184</p><p></p><p>P = 1 / 91</p><p></p><p>∴ Answer: A</p>
Correct Answer: A