Probability of Events and Their Complements: Independence and Conditional Probability

a) To show that P(A|B) = 1 - P(A^c|B), we can start by using the definition of conditional probability:

P(A|B) = P(A ∩ B) / P(B)

Similarly,

P(A^c|B) = P(A^c ∩ B) / P(B)

Since A and A^c are complements, we can use the Law of Total Probability to express P(A) in terms of A and A^c:

P(A) = P(A ∩ B) + P(A ∩ B^c)

Applying this to the numerator of P(A|B), we have:

P(A ∩ B) = P(A)

Substituting this into the expression for P(A|B), we get:

P(A|B) = P(A) / P(B)

Using the complements of A and A^c, we can also express P(A^c) as:

P(A^c) = P(A^c ∩ B) + P(A^c ∩ B^c)

Again, substituting this into the numerator of P(A^c|B), we have:

P(A^c ∩ B) = P(A^c)

Substituting this into the expression for P(A^c|B), we get:

P(A^c|B) = P(A^c) / P(B)

Now, since P(A) + P(A^c) = 1, we have:

P(A^c) = 1 - P(A)

Substituting this into the expression for P(A^c|B), we get:

P(A^c|B) = (1 - P(A)) / P(B)

Finally, we can rewrite this as:

P(A^c|B) = 1 - P(A) / P(B)

This shows that P(A|B) = 1 - P(A^c|B).

b) To show that if A and B are independent, then A^c and B^c are also independent, we need to show that:

P(A^c ∩ B^c) = P(A^c) * P(B^c)

Using the definition of independence, we know that:

P(A ∩ B) = P(A) * P(B)

Taking the complement of both sides:

1 - P(A^c ∩ B^c) = (1 - P(A^c)) * (1 - P(B^c))

Expanding the right side:

1 - P(A^c ∩ B^c) = 1 - P(A^c) - P(B^c) + P(A^c) * P(B^c)

Rearranging the equation:

P(A^c ∩ B^c) = P(A^c) + P(B^c) - P(A^c) * P(B^c)

Since P(A^c) = 1 - P(A) and P(B^c) = 1 - P(B), we can substitute these expressions:

P(A^c ∩ B^c) = 1 - P(A) + 1 - P(B) - (1 - P(A)) * (1 - P(B))

Expanding this equation further:

P(A^c ∩ B^c) = 1 - P(A) + 1 - P(B) - (1 - P(A) - P(B) + P(A) * P(B))

Simplifying:

P(A^c ∩ B^c) = P(A) * P(B)

This shows that P(A^c ∩ B^c) = P(A) * P(B), indicating that A^c and B^c are independent.