   Solutions to Practice Problems for  the Lander Math Contest
Module 12, Solutions (36-39)

36)
The number of outcomes is equal to: 56C4×46=16,
895,340

37) First we find the probability of getting exactly 4 heads out
of 5 tosses: P(4H)=P(F)×P(4H/F)+ P(U)×P(4H/U)=
=(6/15)×5×(0.7)^4×(0.3)+(9/15)×5×(1/2^5)=0.23781
where P(F) and P(U) –the probability of picking fair and unfair
coin respectively.
Then, the conditional probability of picking unfair coin if
exactly 4 heads occurred is equal:
P(U/4H) = P(U and 4H) / P(4H) = P(U)×P(4H/U) / P(4H) = =
(6/15)×5×(0.7)^4×(0.3)/0.23781≈0.6058

38) The conditional probability of picking fair coin if exactly 4
heads occurred is equal:
P(F/4H) = P(F and 4H) / P(4H) = P(F)×P(4H/F) / P(4H) =
= (9/15)×5×(1/2^5)/0.23781≈0.3942
Or, notice that events “F/4H” and “U/4H” are complementary
and: P(F/4H)=1 - P(U/4H)

39) Let us count the probability of the “positive” test:
P(T)=P(D)×P(T/D)+P(H)×P(T/H)=
=(1/10000)×0.99+(9999/10000)×0.01=0.010098≈0.01
Now, we apply Bayes’ theorem:
P(D/T)= P(D)×P(T/D)/P(T)=(1/10000)×0.99/0.01≈0.01

For a good discussion of solution to problem 39, see:
http://www.math.hmc.edu/funfacts/ffiles/30002.6.shtml             