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.

  • 4
    $\begingroup$ Actually I think doing a venn diagram approach is a pretty good method for this question. $\endgroup$ – 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 $$\Bbb P(A^c \cap B^c) = 1 - \Bbb P(A) -\Bbb P(B) + \Bbb P(A \cap B).$$

  • $\begingroup$ thank you very much, could you also tell me, if my calculation is ok? $\endgroup$ – Marie.L Apr 23 at 15:15
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ Thank you, this helps a lot. Could you also tell, if my solution is ok? $\endgroup$ – Marie.L Apr 23 at 15:02

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