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 replaced. The probability that the largest value of tickets drawn is 15 is

$\frac{27}{320}$

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 $\frac{15}{20}=\frac{3}{4}.$

∴ Required probability= Probability of drawing one ticket bearing number 15 and three tickets bearing numbers less than or equal to 15

$= {^4C}_1 ×\frac{1}{20}×\left(\frac{3}{4}\right)^3=\frac{27}{320}$