Question

A box contains $100$ tickets numbered $1$ to $100$. If $2$ tickets are drawn successively without replacement from the box, what is the probability that all the tickets bear numbers divisible by $10$?

• A. $0.01$
• B. $0.008$
• C. $0.009$
• D. $0.007$

Answer 1

The correct option is A $0.01$

Given:

100 tickets, and 2 tickets drawn successively.

Hence n = 2, p = $\frac{10}{100}=\frac{1}{10}$, q = $1-\frac{1}{10}=\frac{9}{10}$

$⟹P\left(\text{all divisible by 10}\right){=}^2{C}_2{\left(\frac{9}{10}\right)}^{2-2}{\left(\frac{1}{10}\right)}^2=1\times 1\times {\left(\frac{1}{10}\right)}^2=0.01$