Questions in relation to Probability

Question

Suppose it is given that $$P(A)=\frac{1}{4},\quad P(B)=\frac{1}{5},\quad P(C)=\frac{3}{5},\quad P(A\cup B\cup C)=1,\quad P(A\cap C)=P(B\cap C)=0.$$ Show that $P(A\cap B)=P(A)\cdot P(B)$ and $P(A\cup B)+P(C)=1.$

For this question, should I show it by using this formula?

$$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B \cap C)$$

Also, would anyone mind telling me how to verify this formula?

Suppose that $A$ and $B$ are two independent events and $P(A)\ne0$. Show that $A$ and $B'$ are also independent.

[Hint: Show that $P(B\mid A)+P(B'\mid A)=1$ by using $(B\cap A)\cup(B'\cap A)=A$.]

For this question, how to show $P(B\mid A) + P(B'\mid A) = 1$?

Answer 1

Yes you can do that for the first question.

As for the second question:

$P(B|A)+P(B'|A)$

$=P(A \cap B)/P(A)+P(A \cap B')/P(A)$

$=(P(A \cap B)+P(A \cap B'))/P(A)$

$=P(A)/P(A)$

$=1$

Answer 2

For the first question, you have the formula:

$P(A\cup B\cup C)=P(A)+P(B)+P(C ) -P(A\cap B)-P(A\cap C)-P(B\cap C) +P(A\cap B\cap C).$

Now just plug in:

$1=\frac{1}{4}+\frac{1}{5}+\frac{3}{5}-P(A\cap B)-0-0+P(A\cap B\cap C)$

$\implies 1=\frac{21}{20}-P(A\cap B)+P(A\cap B\cap C)$

Note at this point that if $P(A\cap C)=P(B\cap C)=0,$ then $P(A\cap B\cap C)=0$ as well. Do you see why? If A and C have no elements in common, and B and C have no elements in common, then there's no way A, B and C can all have some element in common.

So we have: $1=\frac{21}{20}-P(A\cap B)$

$\implies P(A\cap B) = \frac{1}{20}=\frac{1}{4}\frac{1}{5}=P(A)P(B)$

Does that make sense? Can you do other one using a similar method? If you more need help, please let me know in the comments.