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

Options

• \[\frac{5}{17}\]

• \[\frac{4}{17}\]

• \[\frac{5}{16}\]

• \[\frac{5}{18}\]

Answer 1

\[\frac{5}{18}\]

Explanation:

Number of tickets = 9

Number of odd-numbered tickets = 5

Number of even numbered tickets = 4

Required probability

= P {odd, even, odd} + P{even, odd, even}

\[=\frac{^5C_1}{^9C_1}\times\frac{^4C_1}{^8C_1}\times\frac{^4C_1}{^7C_1}+\frac{^4C_1}{^9C_1}\times\frac{^5C_1}{^8C_1}\times\frac{^3C_1}{^7C_1}\]

\[=\frac{5}{9}\times\frac{4}{8}\times\frac{4}{7}+\frac{4}{9}\times\frac{5}{8}\times\frac{3}{7}\]

\[=\frac{80}{504}+\frac{60}{504}\]

\[=\frac{140}{504}=\frac{5}{18}\]