A fair coin is tossed 8 times,what is the probability ofgetting:

1.exactly 3 heads

2.at least one head

3.at most two heads(using binomial distribution)

Solution:

(1) P(exactly 3 heads) = C(8,3)*(1/2)^8 = 0.21875

(2) P(at least one head) = 1- P(no heads) = 1- 0.5^8 = 1- 0.0039 = 0.9961

(3) P(at most two heads) = P(no heads) + P(one head) + P(two heads)

    =  C(8,0)*(1/2)^8 +  C(8,1)*(1/2)^8 +  C(8,2)*(1/2)^8

    = 0.0039 + 0.03125 + 0.109375

    = 0.1445