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एक बैग में `1` से `30` तक अंकित कुल `30` टिकट है । इनमे से यादृच्छया पांच टिकट खींचे जाते है तथा इनको अंकित संख्या के बढ़ते क्रम में रखते है । इनमे से तीसरे टिकट पर संख्या `20` हो, इसकी प्रायिकता होगी
A. `(.^(19)C_(2)xx^(9)C_(2))/(.^(30)C_(5))`
B. `(.^(20)C_(2)xx^(10)C_(2))/(.^(30)C_(5))`
C. `(.^(19)C_(2)xx^(10)C_(2))/(.^(30)C_(5))`
D. `(.^(20)C_(2)xx^(9)C_(2))/(.^(30)C_(5))`

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Correct Answer - C
`30` टिकटो में से `5` खींचने के प्रकार `=^(30)C_(5)`, यदि हम प्रथम `19` टिकटो (`1` से `19` तक ) में से दो टिकट तथा अंतिम `10` टिकटों (`21` से `30` तक ) में से दो टिकट खींचे तथा इनको बढ़ते क्रम में रखे , तो तीसरे टिकट की संख्या `20` होगी|
अतः अभीष्ट प्रायिकता `=(.^(19)C_(2)xx^(10)C_(2))/(.^(30)C_(5))`

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