Suppose that a fair coin with p 5 probability of heads is t

Suppose that a fair coin with p = .5 (probability of heads) is tossed independently 10 times. Find the probability that strictly more heads are obtained than tails.

Solution

Probability of a head(p) = 0.5

Probability of a tail(q) = 0.5

Required Probability = 10C0*(0.5)^10*(0.5)^0 + 10C1(0.5)^9*(0.5) + 10C2(0.5)^8*(0.5)^2 + 10C3(0.5)^7*(0.5)^3 + 10C4(0.5)^6*(0.5)^4

=> P = 0.5^10*[10C0 + 10C1 + 10C2 + 10C3 + 10C4]

=> P = 0.5^10*[1+10+45+120+210]

=> P = 0.37695