I'm preparing for a statistics exam and stumbled upon the following exercisequestion:
In the limited probability space $(\Omega, P)$ let A and B be events. Given $P(A) \ge 0.99$ and $P(B) \ge 0.97$ show that $P(A\cap B) \ge 0.96$.
Now the solution says the following:
$P(A\cap B)= P(A)+P(B)-P(A\cup B)$
$\geq0.99+0.97 -1 = 0.96$
Now my question is: Why is $P(A\cup B)$ equal to one? Isn't there a chance of $A^c$ and $B^c$ equal to $0.99*0.97$?
For a given probability space: Why is the union of A and B equal to 1?
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probability-theory
statistics
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3The reasoning is not that $P(A\cup B)=1$ but that $P(A\cup B)\leqslant P(\Omega)=1$ hence $$P(A\cap B)=P(A)+P(B)-P(A\cup B)\geqslant P(A)+P(B)-1$$ – 2017-01-22
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0@Did alright, I got it. Thank you very much :) – 2017-01-22