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MHT CET · Maths · Probability

A box contains 9 tickets numbered 1 to 9 both inclusive. If 3 tickets are drawn from the box one at a time, then the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

  1. A \(\frac{5}{17}\)
  2. B \(\frac{4}{17}\)
  3. C \(\frac{5}{16}\)
  4. D \(\frac{5}{18}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{5}{18}\)

Step-by-step Solution

Detailed explanation

Odd numbers: 5; Even numbers: 4 Total ways: \(P(9,3) = 9 \times 8 \times 7 = 504\)