It is an example of discrete probability as no of runs will be an integral value ranging from 1 to 3..
So P(1 run) = P(TTT) + P(HHH)
= 1 / 4
P(2 runs) = P(HHT) + P(THH) + P(TTH) + P(HTT)
= 2 / 4
P(3 runs) = P(HTH) + P(THT)
= 1 / 4
Therefore expected number of runs , meaning
E(run) = Σ (no of runs) * P(no of runs)
= (1 * P(no of runs = 1)) + (2 * P(no of runs = 2)) + (3 * P(no of runs = 3))
= (1 * 1 / 4) + (2 * 2 / 4) + (3 * 1 / 4)
= (1 / 4) + (4 / 4) + (3 / 4)
= 8 / 4
= 2
Hence , expected no of runs = 2