Mathematics - Probability Question with Solution | TestHub

In any trail, P (getting white ball) $=\frac{1}{2}$

P (getting black ball) $\begin{array}{rlr}& =1& 2\end{array}$

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 ${}^6{C}_3\times {}^3{C}_3=20_{\text{ways. }}$

So, required probability $=20\times {\left(\frac{1}{2}\right)}^3\times {\left(\frac{1}{2}\right)}^3\times \frac{1}{2}=\frac{5}{32}$