36)

895,340

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

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)

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

http://www.math.hmc.edu/funfacts/ffiles/30002.6.shtml