Question

In a lottery were $90$ tickets numbered $1$ to $90$. Five tickets were drawn at random. The probability that two of the tickets draw numbers $15$ and $89$ is

• A. $\frac{2}{801}$
• B. $\frac{2}{623}$
• C. $\frac{1}{267}$
• D. $\frac{1}{623}$

Answer 1

The correct option is A

$\frac{2}{801}$

Explanation for the correct answer:

Find the probability.

Given,

Five tickets were drawn at random.

So, the total number of ways $={}^{90}{C}_5$

The total number of favorable cases if two of the tickets draw numbers $15$ and $89$ are ${}^{88}{C}_3$

Therefore, the required probability is

$\begin{array}{rcl}\frac{{}^{88}{C}_3}{{}^{90}{C}_5}& =& \frac{{\displaystyle \frac{88\times 87\times 86}{3!}}}{{\displaystyle \frac{90\times 89\times 88\times 87\times 86}{5!}}}\mathbf{[}{\mathbf{\because }}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}\mathbf{]}\\ & =& \frac{2}{801}\end{array}$

Hence, the correct option is A.