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प्रश्न
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
विकल्प
\[\frac{5}{17}\]
\[\frac{4}{17}\]
\[\frac{5}{16}\]
\[\frac{5}{18}\]
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उत्तर
\[\frac{5}{18}\]
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}
\[=\frac{^5C_1}{^9C_1}\times\frac{^4C_1}{^8C_1}\times\frac{^4C_1}{^7C_1}+\frac{^4C_1}{^9C_1}\times\frac{^5C_1}{^8C_1}\times\frac{^3C_1}{^7C_1}\]
\[=\frac{5}{9}\times\frac{4}{8}\times\frac{4}{7}+\frac{4}{9}\times\frac{5}{8}\times\frac{3}{7}\]
\[=\frac{80}{504}+\frac{60}{504}\]
\[=\frac{140}{504}=\frac{5}{18}\]
