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Let's say in an exam, there are two exam takers $A_1$ and $A_2$, each of them takes the exam once.

The exam has two outcomes, Pass (p) and Fail (f). So, the sample space $\Omega=\{p,f\}$.

Further suppose that the probability that $A_1$ passes is 0.6, whereas $A_2$ wins is 0.4, and that the probabilities of winning of the two players are independent.

Let random variables $X_1=1$ if $P_1$ passes and $=0$ otherwise. $X_2$ is defined in the same manner. Then, we have $Pr(X_1=1)=Pr(\{p\})$ and $Pr(X_2=1)=Pr(\{p\})$, suggesting that they should have the same probability of passing, but they do not!

I wonder what is wrong?

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    The problem is that your probability space is way too small to encode all of the information in the problem. Your space only takes into account pass or failure, without considering the fact that there are two independent players here. So, you'd need a 4-point space. – 2017-02-13
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    Well you haven't defined $\Omega$ correctly, that's all. There are 4 outcomes : $A_1$ and $A_2$ pass, $A_1$ passes and $A_2$ doesn't, etc. – 2017-02-13
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    Thank you all very much! You guys are awesome! – 2017-02-13

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The sample space is not $\{p.f\}.$ It is $\{(p,p),(p,f),(f,p),(f,f)\}$ where the first coordinate indcates whether person one passed and the second indicates whether person two passed.

$P(X_1=1) = P(\{(p,p), (p,f)\})$ and $P(X_2 = 1) = P(\{(p,p), (f,p)\}).$

Since they are independent $$P(\{(p,p)\}) = 0.6*0.4=0.24, \\P(\{(p,f)\}) = 0.6*0.6 = 0.36,\\P(\{(f,p)\}) = 0.4*0.4 = 0.16,\\P(\{(f,f)\}) = 0.4*0.6 = 0.24.$$ Now the probability space is fully specified and we can compute $P(X_1=1) = 0.24+0.36 =0.6$ and $P(X_2=1) = 0.24+0.16 = 0.4$

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    I see. Thank you very much for your detailed answer!!! – 2017-02-13