Answer:
0.625
Step-by-step explanation:
Given that {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it applies to the partition {A1, A2} and the event B.
Since A1 and A2 are mutually exclusive and exhaustive, we can say
b) P(B) = P(A1B)+P(A2B)
Selecting any one coin is having probability 0.50. and A1, A2 are events that the coins show heads.[tex]P(B/A1) = 0.50 \\P(B/A2) = 0.75\\P(A1B) = 0.5(0.5) = 0.25 \\P(A2B) = 0.75(0.5) = 0.375\\P(B) = 0.625[/tex]
c) Using Bayes theorem
conditional probability that it wasthe biased coin
=[tex]\frac{0.375}{0.625} \\=\frac{3}{5}[/tex]
d) Given that the chosen coin flips tails,the conditional probability that it was the biased coin=[tex]\frac{0.25*0.5}{0.25*0.5+0.5*0.5} \\=\frac{1}{3}[/tex]