Let events $A_1$, $A_2$, $A_3$ be three mutually independent events such that $P(A_j)$ = $(0.5)^j$ for $j=1,2,3$.
How to find the $P((A_1 \cup A_2) - A_3)$?
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Did you mean to say $[P(A_1 \cup A_2)-P(A_3)]$? – SchrodingersCat Oct 24 '15 at 12:05
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@Aniket, you are right. I have just edited the $A_3$. Thanks – Ganjira Oct 24 '15 at 12:11
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@Ganjira if you mean by the - set minus , then check the edit . – Nizar Oct 24 '15 at 12:27
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1@Nizar, I don't mean the - set minus. – Ganjira Oct 24 '15 at 12:30
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@Ganjira so this minus refers to what? – Nizar Oct 24 '15 at 12:33
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$$P(A_1\cup A_2)-P(A_3)=P(A_1)+P(A_2)-P(A_1\cap A_2)-P(A_3)$$ $$=P(A_1)+P(A_2)-P(A_1)\cdot P(A_2)-P(A_3)$$ since the events are independent $$=(0.5)^1+(0.5)^2-(0.5)^1 \cdot (0.5)^2+(0.5)^3$$ $$=\frac{3}{4}-\frac{1}{8}+\frac{1}{8}$$ $$=\frac{3}{4}$$ $$$$
SchrodingersCat
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