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Question 147

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

Solution

Consider the first case of Odd-Even-Odd:
$$\frac{5}{9}\cdot\frac{4}{8}\cdot\frac{4}{7}=\frac{20}{126}$$

Consider the second case of Even-Odd-Even:
$$\frac{4}{9}\cdot\frac{5}{8}\cdot\frac{3}{7}=\frac{15}{126}$$

Hence, The net probability = $$\frac{20}{126}+\frac{15}{126}=\frac{35}{126}=\frac{5}{18}$$

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