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.

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