Question 599318
You need 2/4 heads followed by 3/6 heads. The probabilities of these events happening are

*[tex \LARGE P(2/4 heads) = {{4} \choose {2}} (\frac{1}{2})^4]


*[tex \LARGE P(3/6 heads) = {{6} \choose {3}} (\frac{1}{2})^6]


Multiply these to get the answer.