A bag contains tickets numbered 1,2,3..........50 of which 5 tickets x1,x2,x3,x4,x5 are drawn at random and arranged in ascending order of magnitude x1<x2<x3<x4<x5.what is the probability that x3=30?

For x₃ to be = 30, x₁ & x₂ must be less than 30 along with x₄ & x₅ being greater than 30.....

Now, No. of all possible selections = ⁵⁰C₅ = a

No of selections such that x₃ is equal to 30 = ²⁹C₂ * ²⁰C₂ = b

Hence, Probability = b/a = (U calculate it)

Let the third ticket be 30.

So, first two tickets have to be less than 30. There are 29 tickets less than 30 out of which two can be selected in ²⁹C₂ ways. Third ticket 30 can be selected in 1 way only. The remaining two are to be selected from 20 tickets numbered greater than 30 which can be selected in ²⁰C₂ ways.

Five tickets can be selected from 50 tickets in ⁵⁰C₅ ways.

So, req. probability = ²⁹C₂ X 1 X ²⁰C₂ / ⁵⁰C₂ = 551 / 15134 (Ans)

@Deepak: U've done a little typing mistake in the last line....................Its ⁵⁰C₅ in the denominator, not ⁵⁰C₂...........