# Question

Suppose we know P(A),P(B | A), and P(Bc | Ac), but we want to find P(A | B).

a. Using the definition of conditional probability for P(A | B) and for P(B | A), explain why P(A | B) = P(A and B)/P(B) = [P(A)P(B | A)]/P(B).

b. Splitting the event that B occurs into two parts, according to whether or not A occurs, explain why P(B) = P(B and A) + P(B and Ac).

c. Using part b and the definition of conditional probability, explain why

P(B) = P(A)P(B | A) + P(Ac)P(B | Ac).

d. Combining what you have shown in parts a–c, reason that

P(A | B) =

P(A)P(B |A)

P(A)P(B | A) + P(Ac)P(B | Ac).

This formula is called Bayes’s rule. It is named after a British clergyman who discovered the

formula in 1763.

a. Using the definition of conditional probability for P(A | B) and for P(B | A), explain why P(A | B) = P(A and B)/P(B) = [P(A)P(B | A)]/P(B).

b. Splitting the event that B occurs into two parts, according to whether or not A occurs, explain why P(B) = P(B and A) + P(B and Ac).

c. Using part b and the definition of conditional probability, explain why

P(B) = P(A)P(B | A) + P(Ac)P(B | Ac).

d. Combining what you have shown in parts a–c, reason that

P(A | B) =

P(A)P(B |A)

P(A)P(B | A) + P(Ac)P(B | Ac).

This formula is called Bayes’s rule. It is named after a British clergyman who discovered the

formula in 1763.

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