# Two independent events, $A$ and $B$; find $B$ given only $A$ and $A\cup B$

Sorry to be a burden but my maths exam is tomorrow and I have no idea how to do this at all. Here is the exact question :

The two independent events $A$ and $B$ are such that $P(A) = 0.2$, $P(A\cup B) = 0.4$.

• a. Evaluate $P(B)$.
• b. Find the probability that at least one of the two events occurs.
• c. Given that exactly one of the events occurs, find the probability that $A$ occurs.

I need help with the entire question really, but at least part a.

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What is the definition of independent events? What is $P(A \cup B)$ in terms of $P(A)$, $P(B)$, etc? Is it true that $x - xy = x(1-y)$ where $x$ and $y$ are real numbers? – Dilip Sarwate Dec 12 '11 at 20:40
(b) is $P(A \cup B)$. (c) is $\dfrac{P(A)-P(A \cap B)}{P(A \cup B) -P(A \cap B)}$ – Henry Dec 12 '11 at 22:52
@Dilip. I think not. It could be $P(A) + P(B) - 2P(A\cap B)$. – Henry Dec 12 '11 at 23:08

Note that $P(A \cup B) = P(A)+P(B)-P(A \cap B)$ where $P(A \cap B) = P(A) \cdot P(B)$.

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But if I don't know what P(ANB) is, how do I work it out? I'm probably missing something really simple here, but I simply don't understand :( – Paul Race Dec 12 '11 at 20:56
@PaulRace: $0.4 = 0.2+P(B)-0.2 \cdot P(B)$. Solve for $P(B)$. – Robert Dec 12 '11 at 21:04

(a) The fact that $$P(A \cup B)=P(A)+P(B)-P(A\cap B)$$ is useful, and the fact that in this case, by independence, we have $P(A \cap B)=P(A)P(B)$ Let $P(B)=y$. Then $$0.4=0.2+y -(0.2)y.$$ The above equation is linear in $y$, and easy to solve. We get $y=0.25$.

(b) A peculiar question! This asks for $P(A \cup B)$, but we have been told that this is $0.4$.

(c) You are probably expected to use the defining formula for conditional probability, namely that $$P(C|D)=\frac{P(C\cap D)}{P(D)}. \qquad(\ast)$$ In our problem, $C$ is the event $A$, and $D$ is the event "exactly one of $A$ and $B$."

First we find $P(D)$. The probability that exactly one of $A$ and $B$ occurs is $P(A \cup B)$ minus the probability that $A$ and $B$ both occur. But since $P(A)=0.2$ and $P(B)=0.25$, by independence $P(A \cap B)=(0.2)(0.25)=0.05$. It follows that the probability that exactly one of $A$ and $B$ occurs is $0.4-0.05$, which is $0.35$.

Now we need $P(C \cap D)$, the probability that $A$ occurs and exactly one of $A$ and $B$ occurs. This is a long-winded way of saying that $A$ occurs and $B$ does not. We have $P(A)=0.2$. By our earlier computation, $P(B)=0.25$, so $P(B^c)$ (the probability $B$ does not occur) is $0.75$. By independence, $P(A \cap B^c)=(0.2)(0.75)=0.15$.

Substituting in $(\ast)$, we find that the probability that $A$ occurs given that exactly one of $A$ and $B$ occurs is $\dfrac{0.15}{0.35}$, which looks better as the fraction $2/7$.

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