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Three distinct numbers a, b and c are chosen at random from the numbers 1, 2, …, 100. The probability that{:("List I","List II"),("a.a, b, c are in AP is","p. "(53)/(161700)),("b.a, b, c are in GP is","q. "(1)/(66)),("c. "(1)/(a). (1)/(b). (1)/(c) " are in GP is ","r. "(1)/(22)),("d.a + b + c is divisible by 2 is","s. "(1)/(2)):} |
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Answer» `{:("a","B","c","d"),("q","s","s","R"):}` a. 2b = a + c = even This means that a and c are both even or both odd. `n(E) = .^(50)C_(2) + .^(50)C_(2) = 50 XX 49` b. Taking r = 2, 3, …, 10, a, b, c can be in GP in 53 ways. c.`(1)/(a), (1)/(b), (1)/(c)` are in GP = a, b, c are in GP d. P(a + b + c is even) = `((.^(50)C_(3)+.^(50)C_(1) xx .^(50)C_(2))/(.^(100)C_(3))) = (1)/(2)` |
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