Calculating the probability of two events.

Question

I have a problem that asks a few questions on the probability of two events where P(A) = 0.6, P(B) = 0.7, and $P(A \cap B)$ = 0.4

i.) The first question asks to calculate the probability of P(A'|B).

Since $$P(A | B) = \frac{P(A \cap B)}{P(B)}$$

My solution was to do $$P(A' | B) = \frac{P(A' \cap B)}{P(B)} = \frac{(0.4) (0.7)}{0.7} = 0.4$$

However, the answer appeared to be: $$P(A' | B) = \frac{0.4}{0.7} = 0.5714$$

ii.) The second asks to calculate the probability of P(A'|B').

My solution was to do $$P(A'| B') = \frac{P(A' \cap B')}{P(B')} = \frac{(0.4) (0.3)}{0.3} = 0.4$$

However, the answer appeared to be: $$P(A'| B') = \frac{0.1}{0.3} = 0.3333$$

I was wondering if anyone can help me see where I went wrong here with both of these questions, or if I'm misunderstanding the actual formulas and how to use them.

Answer 1

Your task is to use the algebra of sets and rules of probability to express what you seek in terms of the three probabilities that were provided: $\mathsf P(A\cap B)=0.4, \mathsf P(A)=0.6,$ and $\mathsf P(B)=0.7$.

Also clearly $\mathsf P(A\cap B)\neq \mathsf P(A)\,\mathsf P(B)$, so the events are not independent.

Use that $\qquad\mathsf P(A'\cap B) ~{= \mathsf P(B\smallsetminus (B\cap A)) \\ = \mathsf P(B)-\mathsf P(A\cap B)}$

Likewise that $\mathsf P(A'\cap B') ~{= 1-\mathsf P(A\cup B)\\ = 1- \mathsf P(A)-\mathsf P(B)+\mathsf P(A\cap B)}$