There are 12 white and 112 red balls in a bag. Balls are drawn one by one with replacement from the bag. The probability that 7th drawn ball is 4th white, is

$\frac{5}{32}$

We have,

p = probability of getting a white ball in a draw $=\frac{1}{2}$

q = probability of getting a red ball in a draw $=\frac{1}{2}$

Getting 4th white ball in 7th draw means that, we must get 3 white balls in first six draws and a white ball in 7th draw.

∴ Required probability $= \begin{Bmatrix}{^6C}_3 \left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{6-3}\end{Bmatrix} \frac{1}{2}=\frac{5}{32}$