Find
(a) $P\{A \cup B\}$
(b) $P\{A^c\}$
(c) $P\{A^c \cap B\}$
This is what I have right now:
(a) $P\{A \cup B\}=0.4+0.5=0.90$
(b) $P\{A^c\}= 1-0.4=0.60$
(c) $P\{A^c \cap B\}= (0.6)\cdot(0.5)=0.30$
Am I doing it correctly?
Suppose A & B are disjoint events with probability $P\{A\}= 0.4$, $P\{B\}=0.5$
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probability
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0Does "disjoint" mean "mutually exclusive" or "independent"? – 2017-02-08
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0It would be independent – 2017-02-08
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1Then you should be wrong: $P(A\cap B)$ should be $0.4×0.5=0.2$. The rest you can work out yourself. – 2017-02-08
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0Oh okay I see my mistake, thanks! – 2017-02-08
2 Answers
1
According to your comment, $A$ and $B$ are indepentent (not disjoint). Be careful, the two things are not equivalent.
Anyway, if $A$ and $B$ are independent, your answer to (a) is not correct. Indeed, $$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$ Since $A$ and $B$ are independent $P(A\cap B)=P(A)P(B)$. Hence $P(A\cup B)=0.4+0.5-0.2=0.7$. On the other hand, (b) and (c) are correct.
1
For any two events $A$ and $B$, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
If $A$ and $B$ are mutually exclusive, then $P(A\cap B)=0$ and thus $$P(A\cup B)=P(A)+P(B)$$
If $A$ and $B$ are independent, then $P(A\cap B)=P(A)\times P(B)$ and thus $$P(A\cup B)=P(A)+P(B)-P(A)P(B)$$
Thus (a) is wrong, while (b) and (c) are correct.