Question:
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 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
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Without replacement â probabilities multiply stepwise.
Updated On: Apr 26, 2026
- $\frac{5}{17}$
- $\frac{4}{17}$
- $\frac{5}{16}$
- $\frac{5}{18}$
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The Correct Option is D
Solution and Explanation
Concept:
Probability = favourable outcomes / total outcomes. Step 1: Count odds and evens. Odds = 5, Evens = 4
Step 2: Find probability. \[ P(OEO) = \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{4}{7} \] \[ P(EOE) = \frac{4}{9} \cdot \frac{5}{8} \cdot \frac{3}{7} \]
Step 3: Add. \[ = \frac{80 + 60}{504} = \frac{140}{504} = \frac{5}{18} \]
Step 4: Conclusion. Answer = $\frac{5}{18}$
Probability = favourable outcomes / total outcomes. Step 1: Count odds and evens. Odds = 5, Evens = 4
Step 2: Find probability. \[ P(OEO) = \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{4}{7} \] \[ P(EOE) = \frac{4}{9} \cdot \frac{5}{8} \cdot \frac{3}{7} \]
Step 3: Add. \[ = \frac{80 + 60}{504} = \frac{140}{504} = \frac{5}{18} \]
Step 4: Conclusion. Answer = $\frac{5}{18}$
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