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Ok, here is a GRE question I cannot seem to solve. My coworker and I keep getting a negative number.

Of three possible events, event A is independent of the other two, and events B and C are mutually exclusive. The probabilities that the individual events A, B, and C will occur are 0.5, 0.3, and 0.2, respectively. What is the probability that both event A and event C will occur?

Source: ManhattanGREPrep

Can you please help? Thank you.

Edit: Thanks for your answers. The original wording was the following:

Of three possible events, events A and B are independent, and events B and C are mutually exclusive. The probabilities that the individual events A, B, and C will occur are 0.5, 0.3, and 0.2, respectively. What is the probability that both event A and event C will occur?

I believe they had realized the original question had a problem and fixed it. I was not aware of it when I posted this question.

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  • $\begingroup$ 17 minutes. $ $ $\endgroup$ – Did Sep 11 '12 at 22:01
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Independence means that $$P(A|C)=\frac{P(A\cap C)}{P(C)}=P(A)$$ $$\implies P(A\cap C)=P(A)P(C)$$

I.e, the probability of $A$ occurring wasn't affected by the prior occurrence of $C$. So the probability that $A$ and $C$ will occur (which is $P(A\cap C)$) is $$P(A)P(C)=0.2\cdot 0.5=0.1$$

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  • $\begingroup$ manhattanprep.com/gre/ChallengeProblems/LastWeek actually, the answer is claimed to be 0.07 $\endgroup$ – xorxorxor Sep 20 '12 at 17:03
  • $\begingroup$ @xorxorxor The answer they give is incorrect. Let $A$ be a coin coming up heads, $B$ be a random number from $1$ to $10$ being in the range $1,2,3$, and $C$ be that same number being $4,5$. Then $P(A\cap C)=1/2\cdot2/10=1/10$. They said it in the question: $B$ is not an independent choice, so including it again in your answer when you've already indicated that $C$ happens (and so $B$ not happening is redundant) is incorrect. $\endgroup$ – Robert Mastragostino Sep 20 '12 at 18:16
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A is indipendent of C, thus the probability of 'A and C' should simply be the product of the probabilities, i.e. 0.1.

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Since $B$ and $C$ are exclusive, $B$ can not occur if $A$ and $C$ did. But this is irrelevant, since by independence $$ \mathbb{P}\left(A \land C\right) \stackrel{\text{independence}}{=} \mathbb{P}(A) \cdot \mathbb{P}(C) = \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10} $$

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P(AUC)=P(A)+P(C)-P(A∩C) Now A and C are independent, so P(A∩C)=P(A) P(C) Therefore the P(AUC) is 0.5 + 0.2 -(0.5) (0.2)=0.5+0.2-0.1=0.6. Now P(A∩C)=P(A) P(C)=0.1. The problem can only be answered if A and C are independent. The original wording does not provide enough information to solve.

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