Question

A box contains $100$ tickets numbered $1,2,...,100$. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than $10$. The probability that the minimum number on them is $5$ is

• A. $\frac{1}{9}$
• B. $\frac{2}{9}$
• C. $\frac{3}{9}$
• D. $\frac{4}{9}$

Answer 1

Answer: (A)

$\frac{1}{9}$

We need to find the probability of a number being $5$ given that the maximum number is not more than $10$.Thus, we need to find $P\left(\frac{minimum=5}{maximum\le 10}\right)=\frac{P\left(\right(minimum=5)\cap (maximum\le 10\left)\right)}{P(maximum\le 10)}$

Hence, probability = $\frac{\frac{{}^5{C}_1}{{}^{100}{C}_2}}{\frac{{}^{10}{C}_2}{{}^{100}{C}_2}}=\frac{5}{45}=\frac{1}{9}$