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
- A \(\frac{5}{17}\)
- B \(\frac{4}{17}\)
- C \(\frac{5}{16}\)
- D \(\frac{5}{18}\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{18}\)
Step-by-step Solution
Detailed explanation
Odd numbers: 5; Even numbers: 4 Total ways: \(P(9,3) = 9 \times 8 \times 7 = 504\)
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