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I am not sure if I answered this question right:

Suppose that A and B are 2 independent events such that the probability that neither occurs is 0.1 and that the probability of B is 0.2. Find the probability of A.

Now I know that 2 events are independent if their intersection = $P(A) * P(B)$. The problem tells me neither occurs has a probability of $p = 0.1$. The probability that B doesn't occurs is $1 - 0.2 = 0.8$. So if I multiplied B not occurring by A occurring it should give me : $P(A') * P (B') = 0.1 \mapsto P(A') = \frac{0.1}{0.8} = 0,125$. Now $1 - P (A')$ should equal to $P(A)$ which is 1 - 0.125 = 0.875. Is that correct?

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  • $\begingroup$ It is difficult to quickly read what you have written. Visit this page for information on how to type mathematics on this site. $\endgroup$ – JMoravitz Dec 16 '17 at 2:21
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You already correctly noted the definition of independent events: $P(A\cap B)=P(A)P(B)$, then note that $A,B$ independent implies $A,B^c$ independent and $A^c,B^c$ independent, etc... We are told $P(A^c\cap B^c)=0.1$ and $P(B)=0.2$.

So, $$0.1=P(A^c\cap B^c)=P(A^c)P(B^c)=(1-P(A))(1-P(B))=(1-P(A))(1-0.2)$$

At this point it is just algebraic manipulation to complete.

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Another approach:

$P(A^c \cap B^c) = 0.1$ implies that $$P((A^c \cap B^c)^c) =P(A \cup B) = 1- 0.1 = 0.9$$

Since A and B are independent, we can say: $$P(A \cup B) = P(A) + P(B) - P(A)P(B)$$ Solving for $P(A)$ from here gives the same result

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