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

A (1/2)¹¹

B (9)(1/2)

C (11C2)(1/2)⁹

D (1/2)

Correct Answer: Option A

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

= \(\left(\frac12\right)^2\times\left(\frac12\right)^9=\left(\frac12\right)^{11}\)