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i have following question:suppose we have two independence event whose probability are following: P(A)=0.4 and P(b)=0.7,we are asked to find P(A union B) from probability theory i know that P(A union B)=P(A)+P(B)-P(A interesection B) sure last one is equal zero so it means that result should be P(A)+P(B) but it is more then 1 (1.1)please help me where i am wrong?)

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3 Answers 3

up vote 4 down vote accepted

If the events $A$ and $B$ are independent, then $P(A \cap B) = P(A) P(B)$ and not necessarily $0$.

You are confusing independent with mutually exclusive.

For instance, you toss two coins. What is the probability that both show heads? It is $\frac{1}{2} \times \frac{1}{2}$ isn't it? Note that the coin tosses are independent of each other.

Now you toss only one coin, what is the probability that it shows both heads and tails?

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1  
it is zero because if show me heads for example it is impossible show tails at the same time everything is clear thanks –  dato datuashvili Jul 8 '11 at 7:44
    
@user: Yes, zero is correct. Both events are mutually exclusive. –  Aryabhata Jul 8 '11 at 7:48
    
ok i have understood thanks very much it is a very great site with very great users –  dato datuashvili Jul 8 '11 at 7:58

If $A$ and $B$ are 2 independent events then :

\begin{align*} P(A \cup B) &= P(A) + P(B) - P(A)\cdot P(B) \quad \Bigl[\because \tiny P(A \cap B) = P(A) \cdot P(B) \ \text{for independent events} \Bigr] \\ &= \frac{4}{10} + \frac{7}{10} - \frac{28}{100} \\ &= \frac{110-28}{100} = \frac{82}{100} =0.82 \end{align*}

Please refer this link:

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No, the last one is not $0$. If $A,B$ are independent, then $P(A \cap B) = P(A) P(B) = 0.4 \cdot 0.7 = 0.28$

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