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Two number `aa n db`aer chosen at random from the set of first 30 naturalnumbers. Find the probability that `a^2-b^2`is divisible by 3. |
Answer» Correct Answer - `(47)/(87)` The total number of ways of choosing two numbers out of 1, 2, 3, …, 30 is `.^(30)C_(2)` = 435. Now, 30 natural numbers are categorized as 3k type: 3, 6, 9, … 30 `rarr` 10 numbers 3k + 1 type: 1, 4, 7, …, 28 `rarr` 10 numbers 3k + 2 type: 2, 5, 8, ..., 29 `rarr` 10 numbers `a^(2) - b^(2)` is divisible by 3 iff either both of a and b are So, number of favourable cases = `.^(10)C_(2) + .^(20)C_(2) = 235` `therefore` Required probability = `(235)/(435) = (47)/(87)` |
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