If $P(B)>0 , B\subset A$ and $A$ and $B$ are independent, then $P(A)=1$

Question

Let $(\Omega,{\scr F},P)$ be probability space, and $A,B \in {\scr F}$, then:

$(a)$ if $P(B)>0 , B\subset A$ and $A,B$ are independent then $P(A)=1$ ;

$(b)$ if $P(A \cap B)=\frac19$ and $P(A \cup B)=\frac17$ then $P(A \cup B\;|\; A\cap B)=\frac17$ ;

$P(A \cup B\;|\; A\cap B)=\frac{P((A \cup B)\;\cap\; (A\cap B))}{P(A \cap B)} = \frac{P(A \cap B)}{P(A \cap B)}=1\neq\frac17$ ; am I right?

Answer 1

An alternative is to work from the following definition:

$A$ and $B$ are independent iff $P(A|B)=P(A)$ and $P(B|A)=P(B)$.

Here $P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}$. Since $P(A)>0$, then $$P(A|B)=P(A) \quad \iff \quad \frac{P(A)}{P(B)}=P(A) \quad \iff \quad P(B)=1.$$

And is immediate that also $P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{P(A)}{P(A)}=1=P(B)$.

Answer 2

Observe that $A\cap B=B$ so that in fact:$$P(B)=P(A\cap B)=P(A)P(B)$$Now draw conclusions on base of $P(B)>0$.

Answer 3

If $B\subset A$ then $P(A\cap B)=P(B)$. You got that $P(A\cap B)=P(A)P(B)$, so $P(A)P(B)=P(B)$. Because $P(B)>0$ we got $P(A)=1$. Am i right?

Answer 4

Observe that if $A$ and $B$ are independent, then we have that $A^{c}$ and $B$ are independent. Thus $$P(A^{c} \cap B) = P(A^{c})P(B)= (1-P(A))P(B).$$ But $P(A^{c} \cap B)=P(\emptyset)=0$ and since $P(B) > 0$, we must have that $P(A)=1$.