Answer
Three mutually exclusive events can be
E_{1} = Getting exactly one head
E_{2} = Getting exactly two heads
E_{3} = Getting exactly three heads
P(E_{1}) = {HTT, THT, TTH} = 3/8
P(E_{2}) = {HHT, HTH, THH} = 3/8
P(E_{3}) = {HHH} = 1/8
P(A) = P(E_{1}) + P(E_{2}) + P(E_{3})
= 3/8 + 3/8 + 1/8
= 7/8
Work Step by Step
As given in answer
