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A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.
550/15379
550/15134
551 / 15134
552/15379
Total number of elementary events = 50C5 Given,third ticket =30
=> first and second should come from tickets numbered 1 to 29 = 29C2
ways and remaining two in 20C2 ways.
Therfore,favourable number of events = 29C2*20C2
Hence,required probability = 29C2*20C2/50C5 =551 / 15134
Hence, option 3 is the correct answer.
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