A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½, 

⇒ Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads = $\frac{1}{2}2\times \frac{1}{2}9=\frac{1}{2}11$