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This is a question set in my maths class and I think it has a trick question. It's super simple, probably far too simple for this website, but here goes anyway.

Assume that we have two events, $A$ and $B$, that are independent and mutually exclusive. Assume further that we have $P(A)=0.1$ and $P(B)=0.1$. What is $P(A \cap B)$?

I am thinking the answer is $0$, as if the events are mutually exclusive they should never happen at the same time.

The reason I am stuck is the teacher has said the answer is $0.1 \times 0.1$.

Another small point:

Question

If I have a fair coin and I toss it 3 times then I think the sample space is?

{HHH, HHT, HTT, HTH, TTT, TTH, THH, THT}?

Am I correct and if not why not?

I do think they are correct but I would just like to double check?

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    $\begingroup$ it is not possible for two events to be independent and mutually exclusive (excluding cases where one or the other of the events has probability $0$). $\endgroup$ – lulu Feb 25 '19 at 14:05
  • $\begingroup$ what about if I removed the independent part from the question? $\endgroup$ – user22485 Feb 25 '19 at 14:05
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    $\begingroup$ If you just assume they are mutually exclusive then, by definition, $P(A\cap B)=0$. If you just assume they are independent then $P(A\cap B)=P(A)\times P(B)$. $\endgroup$ – lulu Feb 25 '19 at 14:06
  • $\begingroup$ perfect, I should speak to my teacher about this. $\endgroup$ – user22485 Feb 25 '19 at 14:07
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    $\begingroup$ Yes, those are the possible outcomes from three tosses of a coin. $\endgroup$ – lulu Feb 25 '19 at 14:16
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You are right. If two events are mutually exclusive, it means that they cannot happen at the same time (see here). This means that $A\cap B=\varnothing$ and $P(A\cap B)=0$.

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  • $\begingroup$ Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! $\endgroup$ – user22485 Feb 25 '19 at 14:04

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