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Step-by-Step Solution
Step 1: Understand the Problem
We have 11 consecutive natural numbers. We randomly select 3 numbers (without repetition). We want the probability that these chosen 3 numbers are in an arithmetic progression (A.P.) with a positive common difference.
Step 2: Total Number of Ways to Select 3 Numbers
The total number of ways to choose 3 numbers from 11 is given by the combination:
$ ^{11}C_{3} $.
Thus,
$ ^{11}C_{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165.$
Step 3: Condition for Three Numbers to Be in A.P.
If the three numbers are $a, b, c$ (in ascending order), they form an A.P. if
$ 2b = a + c.$
Since we are dealing with consecutive integers, $a + c$ must be even for $b$ to be an integer. This means either $a$ and $c$ must both be even or both be odd.
Step 4: Count the Favorable Selections
Among 11 consecutive natural numbers, there are two possible distributions of evens and odds:
6 evens and 5 odds (e.g., 1 to 11 starting with odd, ending with odd), or
5 evens and 6 odds (e.g., 2 to 12 starting with even, ending with even if we consider another consecutive set).
In each case, to form an A.P., we pick either 2 evens out of the available evens or 2 odds out of the available odds for $a$ and $c$, with $b$ determined uniquely in between.
Step 5: Case-by-Case Probability
Consider the first scenario (6 evens, 5 odds):
Number of ways to choose 2 evens from 6: $ ^{6}C_{2} $
Number of ways to choose 2 odds from 5: $ ^{5}C_{2} $
So, favorable cases $= \, ^{6}C_{2} + \, ^{5}C_{2}.$
The probability in this scenario is:
$ \displaystyle \frac{ ^{6}C_{2} + \, ^{5}C_{2} }{ ^{11}C_{3} }.$
Similarly, for the second scenario (5 evens, 6 odds), the count is the same:
$ \displaystyle \frac{ ^{5}C_{2} + \, ^{6}C_{2} }{ ^{11}C_{3} }.$
Both scenarios are equally likely, and each gives the same probability.
Step 6: Calculate the Probability
First, compute $ ^{6}C_{2} = \frac{6 \times 5}{2} = 15 $ and $ ^{5}C_{2} = \frac{5 \times 4}{2} = 10.$ Thus,
$ ^{6}C_{2} + ^{5}C_{2} = 15 + 10 = 25.$
The probability for one scenario:
$ \displaystyle \frac{25}{ ^{11}C_{3} } \;=\; \frac{25}{165} \;=\; \frac{5}{33}.$
Since both distributions (6 evens, 5 odds) and (5 evens, 6 odds) occur with equal chance, the overall probability remains the same $ \frac{5}{33} $.
Final Answer
The probability that three numbers selected from 11 consecutive natural numbers are in arithmetic progression with a positive common difference is
$ \boxed{\frac{5}{33}}.$
