Probability question - Independent events - find probability of $A$ given $B$

Question

Consider two independent events, $A$ and $B$, where the $P(B)$ is $0.44$ and the probability that $A$ does not occur or $B$ occurs is $0.74$. Determine the probability that event $A$ occurs.

Answer 1

Hint: Use $P(B \text{ does not occurs} | A \text{ occurs})$$

Answer 2

By the law of total probability, $$ P(A)=P(A\cap B)+P(A\cap B^c). $$ Using the independence of $A$ and $B$ and de Morgan's laws, $$ P(A)=P(A)P(B)+1-P(A^c\cup B). $$ Hence, $$ P(A)=0.44P(A)+1-0.74 $$ and $P(A)=0.26/0.56=13/28$.

Answer 3

You have: $0.74 =P(A'\cup B)= P(A') +P(B) - P(A'\cap B)= 1 - P(A) + P(B)- (P(B) - P(A\cap B)) = 1 - P(A) + P(A\cap B) = 1 - P(A) + P(A)P(B)= 1-P(A) + 0.44P(A) = 1 - 0.56P(A)\implies P(A) = \dfrac{0.26}{0.56}= \dfrac{13}{28}$