Question

A bag contains $50$ tickets numbered $1,2,3,\dots \dots ,50$ of which five are drawn at random and arranged in ascending order of magnitude$(X_1<X_2<X_3<X_4<X_5)$. The probability that $X_3=30$ is

• A. $\frac{{}^{20}\mathrm{C}_2}{{}^{50}\mathrm{C}_5}$
• B. $\frac{{}^2\mathrm{C}_2}{{}^{50}\mathrm{C}_5}$
• C. $\frac{{}^{20}\mathrm{C}_2\times {}^{29}\mathrm{C}_2}{{}^{50}\mathrm{C}_5}$
• D. None of these

Answer 1

The correct option is C

$\frac{{}^{20}\mathrm{C}_2\times {}^{29}\mathrm{C}_2}{{}^{50}\mathrm{C}_5}$

Explanation for the correct answer:

Step 1: Express the given data:

The tickets are numbered from $\{1,2,3,.....,50\}$

Total number of tickets = $50$

Five tickets are taken and are arranged in ascending order as $(X_1<X_2<X_3<X_4<X_5)$

The third ticket is given as $X_3=30$

So, the first two tickets should be numbered from $1$ to $29$ and the last two tickets are numbered from $31$to $50$.

Step 2: Finding the probability:

Then the number of ways of selection for the first two and last two tickets = ${}^{20}\mathrm{C}_2\times {}^{29}\mathrm{C}_2$

The number of ways of selecting $5$ tickets from $50$ tickets is $50C_5$

Thus the probability that $X_3=30$is

$=\frac{{}^{20}\mathrm{C}_2\times {}^{29}\mathrm{C}_2}{{}^{50}\mathrm{C}_5}$.

Hence, option (C) is the correct answer.