A box contains 24 identical balls of which 12 are white and 12 black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the  $4^{th}$ time on the  $7^{th}$ draw is

In any trail, P(getting white ball) =  $\frac{1}{2}$
P(getting black ball) =  $\frac{1}{2}$
Now, required event will occur if in the first six trails 3 white balls are drawn in any one of the 
3 trails from six. The remaining 3 trails must be kept reserved for black balls. This can happen 
in ${ }^6{C}_3{\times }^3{C}_3=20$ ways.
So, required probability  $=20\times {\left(\frac{1}{2}\right)}^3\times {\left(\frac{1}{2}\right)}^3\times \frac{1}{2}=\frac{5}{32}$