Probability
Counting Principles and Probability
Grade 12
Question:
<p>Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in AP with positive common difference, is</p>
<p>(a) \(\frac{15}{101}\)</p>
<p>(b) \(\frac{5}{101}\)</p>
<p>(c) \(\frac{5}{33}\)</p>
<p>(d) \(\frac{10}{99}\)</p>
Step-by-Step Solution
Key Concept: Three numbers in AP with positive common difference means we need to count pairs of same parity (both even or both odd) from consecutive natural numbers.
<p><strong>Step 1:</strong> We have to select three numbers at random out of 11 consecutive natural numbers such that they are in AP.</p><p><strong>Step 2:</strong> For three numbers to be in AP, two numbers must be either even or odd and third will be selected automatically.</p><p><strong>Step 3:</strong> Among 11 consecutive natural numbers, there are 6 even numbers and 5 odd numbers (or 6 odd and 5 even, depending on the starting number).</p><p><strong>Step 4:</strong> Number of ways to select 3 numbers in AP = $\binom{6}{2} + \binom{5}{2}$</p><p><strong>Step 5:</strong> Total number of ways to select 3 numbers out of 11 is $\binom{11}{3}$</p><p><strong>Step 6:</strong> Probability = $\frac{\binom{6}{2} + \binom{5}{2}}{\binom{11}{3}} = \frac{15 + 10}{165} = \frac{25}{165} = \frac{5}{33}$</p><p>∴ Answer is C.</p>
Correct Answer: C