Solutions to Practice Problems for  the Lander Math Contest
Module 8, Solutions (23-25)

23) The event “three heads” would occur in two ways: first,
choose a fair coin, then get three heads and one tail tossing
the fair coin; and second, choose an unfair coin, then get
three heads and one tail tossing the unfair coin:
P(3H,T)=P(F)*P(3H,T|F)+P(U)*P(3H,T|U)= (8/10)(4/2^4)+
(2/10)(4*0.9^3*0.1) =0.25832

24) Let us apply Bayes’ formula:
P(F|3H,T)= P(F)*P(3H,T|F) / P(3H,T)= (8/10)(4/2^4/025832=
0.7742

25) First we calculate the probability of having exactly two
heads and two tails:
P(2H,2T)=P(F)*P(2H,2T|F)+P(U)*P(2H,2T|U)=(8/10)(6/2^4)+
(2/10)(4*0.9^2*0.1^2) =0.30648
Then, Bayes’ formula immediately gives us the answer:
P(F|2H,2T)= P(F)*P(2H,2T|F) / P(2H,2T)=(8/10)(6/2^4)/
0.30648≈0.6525