<p>Out of 21 tickets consecutively numbered, there are drawn at random. Find the probability that the numbers on them are in AP. If the probability is \(\frac{a}{b}\), then \((14a - b)\) is ……….</p>
Step-by-Step Solution
Key Concept: Count sequences in arithmetic progression by parameterizing by first term and common difference, then use the probability formula.
<p><strong>Solution:</strong></p><p>Total ways to choose 3 tickets from 21: \(\binom{21}{3} = \frac{21 \times 20 \times 19}{6} = 1330\)</p><p>For three numbers in AP with first term \(a\) and common difference \(d\): the numbers are \(a, a+d, a+2d\).</p><p>We need: \(1 \le a < a+d < a+2d \le 21\)</p><p>This gives: \(a + 2d \le 21\), so \(a \le 21 - 2d\)</p><p>For each \(d\), number of valid \(a\) values: \(21 - 2d\)</p><p>Number of favorable outcomes = \(\sum_{d=1}^{10}(21-2d) = 19 + 17 + 15 + ... + 1 = 100\)</p><p>Probability = \(\frac{100}{1330} = \frac{10}{133}\)</p><p>So \(a = 10, b = 133\)</p><p>\(14a - b = 14(10) - 133 = 140 - 133 = 7\)</p>
Correct Answer: 7