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A box contains 10 tickets numbered from 1 to 10 . Three tickets are drawn one by one with replacement. The probability that the number on the first ticket drawn is strictly greater than the number on the second ticket, which in turn is strictly greater than the number on the third ticket, is equal to

Options

  1. A18 25
  2. B1 6
  3. C11 50
  4. D3 25

Correct answer

D. 3 25

Step-by-step solution

The experiment consists of drawing 3 tickets with replacement from a box of 10 tickets. Total number of possible outcomes = 10 10 10 = 1000 . For the sequence of drawn numbers to be strictly decreasing, all three numbers drawn must be distinct. The number of ways to choose 3 distinct numbers from the 10 available numbers is ¹⁰C₃ . Once 3 distinct numbers are chosen, there is exactly 1 way to arrange them in a strictly decreasing order. Number of favourable outcomes = ¹⁰C₃ = 10 9 8 3 2 1 = 120 . Required probability

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