A bag contains 50 tickets numbered 1, 2, 3, .., 50 of which five are drawn at random and arranged in ascending order of magnitude`(x_1ltx_2ltx_3ltx_4ltx_5)` find the probability that `x_3=30`.

Answer

`(.^(29)C_(2)xx.^(20)C_(2))/(.^(50)C_(5))`

Explanation

Five tickets out of 50 can be drawn in `.^(50)C_(5)` ways. Since `x_(1) lt x_(2) lt x_(3)lt x_(4) lt x_(5)` and `x_(3) = 30`, we have `x_(1), x_(2) lt 30`, i.e., `x_(1)` and `x_(2)` should come from tickets numbered 1 and 29 and this may happen in `.^(29)C_(2)`. Remaining ways, i.e., `x_(4), x_(5) gt 30`, should come from 20 tickets numbered 31 to 50 in `.^(20)C_(2)` ways.
So, favorable number of cases is `.^(29)C_(2) .^(20)C_(2)`. Hence, required probability is
`(.^(29)C_(2) xx .^(20)C_(2))/(.^(50)C_(5))`