Question

A bag contains $50$ tickets numbered $1,2,3,...,50$ of which five are drawn at random and arranged in ascending order of magnitude $(x_1<x_2<x_3<x_4<x_5),$ then the probability that $x_3=30$ is

• A. $\frac{{}^{29}{C}_2{\times }^{20}{C}_2}{{}^{50}{C}_5}$
• B. $\frac{{}^{30}{C}_1{\times }^{29}{C}_1}{{}^{50}{C}_5}$
• C. $\frac{{}^5{C}_1{\times }^{50}{C}_2}{{}^{50}{C}_5}$
• D. $\frac{{}^{50}{C}_2{\times }^{29}{C}_1}{{}^{50}{C}_5}$

Answer 1

Answer: (A)

Let E= Event of getting 5 tickets ascending order. Since, it is given that number 30 is on 3rd position. So we have to choose two numbers from 1 to 29 numbers and choose two numbers from 31 to 50
$\therefore$ $n(E){=}^{29}{C}_2{\times }^{20}{C}_2$
Total number of selection of 5 cards From 1 to 50 is,
$n(S){=}^{50}{C}_5$
$\therefore$ Required probability
$=\frac{n(E)}{n(S)}=\frac{{}^{29}{C}_2{\times }^{20}{C}_2}{{}^{50}{C}_5}$