# How could I calculate the probability of $\mathbb{P}\left(A^{c} | B\right)$ and $\mathbb{P}\left(B^{c} | A^{c}\right)$

Good morning! The task is to calculate the probability of $$\mathbb{P}\left(A^{c} | B\right)$$ and $$\mathbb{P}\left(B^{c} | A^{c}\right)$$.

Given are the probabilities of $$\mathbb{P}(A)=2 / 3, \mathbb{P}(B)=1 / 2$$ and $$\mathbb{P}(A \cap B)=2 / 5$$

Possible solution?:

First I use the formula of the conditional probability

(i) $$\mathbb{P}\left(A^{c} | B\right)$$ = $$\frac{\mathbb{P}(A^c \cap B)}{\mathbb{P}(B)}$$

But now I have this problem here $$\mathbb{P}(A^c \cap B)$$

So I used the bad way and draw a circle too become an equation.

(ii) $$\mathbb{P}(A^c \cap B) = \mathbb{P}(B)-\mathbb{P}(A \cap B)$$

Now I can write (i) with (ii) too $$\mathbb{P}\left(A^{c} | B\right)$$ = $$\frac{\mathbb{P}(B)-\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$ = $$\frac{\frac{1}{2}-\frac{2}{5}}{\frac{1}{2}}=\frac{3}{5}$$

Now the second probability: $$\mathbb{P}\left(B^{c} | A^{c}\right)= \frac{\mathbb{P}(A^c \cap B^c)}{\mathbb{P}(B)}$$

With $$\mathbb{P}(A^c \cap B^c) = 1-\mathbb{P}(A)-\mathbb{P}(B)-\mathbb{P}(A \cap B)$$

We have

$$\mathbb{P}\left(B^{c} | A^{c}\right)= \frac{1-\mathbb{P}(A)-\mathbb{P}(B)-\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{1-\frac{2}{3}-\frac{1}{2}+\frac{2}{5}}{\frac{1}{2}} = \frac{7}{15}$$

I know, drawing a circle is a bad way, but I do not know how to proof all the equations.

Could someone help me to solve those probabilities, because I believe my way is wrong.

• Actually I think doing a venn diagram approach is a pretty good method for this question. – George Dewhirst Apr 23 at 14:53

Hint $$:$$ Observe that $$(A \cap B) \cup (A^c \cap B) = B$$ and $$(A \cap B) \cap (A^c \cap B) = \varnothing.$$ So we have $$\Bbb P(A^c \cap B) = \Bbb P(B) - \Bbb P(A \cap B).\ \ \ \ (1)$$
Also $$(A^c \cap B) \cup (A^c \cap B^c) = A^c$$ and also $$(A^c \cap B) \cap (A^c \cap B^c) = \varnothing.$$ So we have $$\Bbb P(A^c \cap B^c) = \Bbb P(A^c) - \Bbb P(A^c \cap B).\ \ \ \ (2)$$ Now we know that $$\Bbb P(A^c)= 1 - \Bbb P(A).$$ So using $$(1)$$ and $$(2)$$ we have $$\Bbb P(A^c \cap B^c) = 1 - \Bbb P(A) -\Bbb P(B) + \Bbb P(A \cap B).$$
• You have written $$\Bbb P(A^c \cap B^c) = 1 - \Bbb P(A) -\Bbb P(B) - \Bbb P(A \cap B).$$ Which is not true. Instead it would be $$\Bbb P(A^c \cap B^c) = 1 - \Bbb P(A) -\Bbb P(B) + \Bbb P(A \cap B)$$ as I have mentioned above. – Dbchatto67 Apr 23 at 15:37
The identities derived from venn diagrams are correct. If you want to be more formal you could note that $$(B\cap A^c)\cup (A\cap B)=B$$ where the union is a disjoint union. It follows that $$P(B)=P(B\cap A^c)+P(A\cap B).$$ To prove the set theoretic identity, you could proceed as follows $$B=B\cap \Omega=B\cap (A\cup A^c)=(B\cap A)\cup (B\cap A^c).$$ I leave it to you to prove that the union is disjoint.