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Let $ A_1, A_2, A_3 $ be independent events with probabilities $ \frac 12,\frac13,\frac14,$ respectively. How to compute for $P(A_1\cup A_2 \cup A_3).$

My solution starts from using the probability of their complements, I do not know how to answer this question. Please help.

$P(A_1\cup A_2 \cup A_3) = 1- P(A_1^c\cap A_2^c \cap A_3^c)$

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    $\begingroup$ Why don't you write down the part you've already done? $\endgroup$
    – user180040
    Oct 4, 2014 at 4:49
  • $\begingroup$ I already edited my solution $\endgroup$ Oct 4, 2014 at 4:57
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    $\begingroup$ That's a good start. Are there any assumptions under which you would be able to calculate $P(A_1^c \cap A_2^c \cap A_3^c)$? $\endgroup$
    – user180040
    Oct 4, 2014 at 4:59
  • $\begingroup$ getting the complements of the events first $\endgroup$ Oct 4, 2014 at 5:00
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    $\begingroup$ You mean $P(A_1^c) = 1/2$, $P(A_2^c) = 2/3$, $P(A_3^c) = 3/4$, right? $\endgroup$
    – user180040
    Oct 4, 2014 at 5:04

2 Answers 2

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$P(A_1\cup A_2 \cup A_3)= P(A_1\cup A_2) + P(A_3) - P((A_1\cup A_2)\cap A_3 )$ $=P(A_1\cup A_2) + P(A_3) - P(A_1\cup A_2).P(A_3 )$ ..since $(A_1\cup A_2)$ and $(A_3)$ are independent

Similarly, $P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=P(A_1)+P(A_2) - P(A_1).P(A_2)$

Now you can just do the calculation. I guess the final answer is 3/4

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    $\begingroup$ First the formula for P(A∪B∪C) is known in full generality. Second this is definitely not the most efficient approach. $\endgroup$
    – Did
    Jul 27, 2015 at 22:07
  • $\begingroup$ @Did - As far as efficient approach is concerned, are you talking about $$P(A_1 \cup A_2 \cup A_3) = 1 - P( (A_1 \cup A_2 \cup A_3)^c) = 1- P(A_1^c \cap A_2^c \cap A_3^c) = 1 - P(A_1^c).P(A_2^c).P(A_3^c)$$ Btw, not sure what do you mean by " First the formula for P(A∪B∪C) is known in full generality " $\endgroup$
    – KGhatak
    Jul 28, 2015 at 15:16
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    $\begingroup$ 1. Yes. 2. That the formula for the probability of a union is known in full generality as the alternated sum of the probabilities of the events, minus the sum of the probabilities of the two-by-two intersections, plus the sum of the probabilities of the three-by-three intersections, etc., except that starting from three events there is no "etc." And also that your answer seems to be left incomplete (one expects that you wuold collect every term to get the full formula for P(A∪B∪C) but you don't...) $\endgroup$
    – Did
    Jul 28, 2015 at 16:55
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Use the following fact

If $A$ and $B$ are independent events, then $A^{c}$ and $B^{c}$ are independent evetns

Proof: We see that $$P(A^c\cap B^c)=P(A^c)+P(B^c)-P(A^c\cup B^c)=P(A^c)+P(B^c)-P((A\cap B)^c)=$$ $$P(A^c)+P(B^c)-(1-P(A\cap B))$$ which since $A$ and $B$ are independent then we have $$=P(A^c)+P(B^c)-(1-P(A)P(B))=(1-P(A))+(1-P(B))-(1-P(A)P(B))=$$ $$1-P(A)-P(B)+P(B)P(A)= (1-P(A))(1-P(B))=P(A^c)P(B^c)$$

Thus $A^c$ and $B^c$ are independent

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    $\begingroup$ This fact, while true, is not enough to solve the exercise. $\endgroup$
    – Did
    Jul 27, 2015 at 22:06
  • $\begingroup$ Well doesn't the definition of independence with this just convert $P(A_1^c \cap A_2^c \cap A_3^c)=P(A_1^c)P(A_2^c)P(A_3^c)$ $\endgroup$
    – Kamster
    Jul 28, 2015 at 1:11
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    $\begingroup$ No, the independence of 3 events is not reducible to 2-by-2 independences. $\endgroup$
    – Did
    Jul 28, 2015 at 5:23
  • $\begingroup$ So it's not true if for three events that each are mutually/ pair wise independent that the 3 events are independent $\endgroup$
    – Kamster
    Jul 28, 2015 at 5:26
  • $\begingroup$ Indeed. $ $ $ $ $\endgroup$
    – Did
    Jul 28, 2015 at 5:27

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