Question

From a box containing $20$ tickets marked with numbers $1$ to $20$, four tickets are drawn one-by-one. After each draw, the ticket is replace. The probability that the largest value of tickets drawn is $15$, is

• A. ${\left(3/4\right)}^4$
• B. $\frac{27}{320}$
• C. $\frac{27}{1280}$
• D. $\frac{9}{16}$

Answer 1

Answer: (B)

$\frac{27}{320}$

We have
Probability of drawing a ticket bearing number $15$ is $\frac{1}{20}$
Probability of drawing a ticket bearing a number less than or equal to $15$ is $\frac{15}{20}=\frac{3}{4}$
$\therefore$ Required Probability =Probability of drawing one ticket bearing number $15$ and three tickets bearing numbers less than or equal to $15$
$={{}^{4}C}_1\times \frac{1}{20}\times {\left(\frac{3}{4}\right)}^3=\frac{27}{320}$