Probability Discrete Math


An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is



option (i) is correct.

Just solve it this way.

E = 2*(1/2) + 3*(1/2^2) + 4*(1/2^3) + … upto infinity

Solving the above AGP we get,