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I have following question:

Suppose we have two independent events whose probability are the following: $P(A)=0.4$ and $P(B)=0.7$.

We are asked to find $P(A \cap B)$ from probability theory. I know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$. But surely the last one is equal zero so it means that result should be $P(A)+P(B)$ but it is more than $1$ (To be exact it is $1.1$). Please help me where i am wrong?

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4 Answers 4

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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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    $\begingroup$ it is zero because if show me heads for example it is impossible show tails at the same time everything is clear thanks $\endgroup$ Jul 8, 2011 at 7:44
  • $\begingroup$ @user: Yes, zero is correct. Both events are mutually exclusive. $\endgroup$
    – Aryabhata
    Jul 8, 2011 at 7:48
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    $\begingroup$ ok i have understood thanks very much it is a very great site with very great users $\endgroup$ Jul 8, 2011 at 7:58
  • $\begingroup$ But tossing 2 coins looks like a union: we're asking for probability of 2 independent events at the same time. I'd even suppose that intersection here is equal to union, but their formulas would give different results. I am so confused. $\endgroup$
    – Hi-Angel
    Apr 17, 2017 at 23:10
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    $\begingroup$ @Hi-Angel: Suppose you toss two coins (call them A and B) at the same time. What is probability that at least one of them shows heads? This is asking about the union. What is the probability that both show heads? This is asking about the intersection. $\endgroup$
    – Aryabhata
    Apr 17, 2017 at 23:18
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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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All things considered, it is important to realize the great difference between mutually exclusive and independent events. This is what throws everybody off in my math class. Mutually exclusive is when two events cannot happen at the same time and independent is when two events do not influence each other. In this case, there is an intersection because there are independent events, so you would subtract 1/28. On the other hand, let's look at it this way. If A was the chance of picking a king (4 kings) from a deck of cards and B was the chance of picking a queen as a deck of cards (4 queens), what is the probability that you would pick a king or a queen. Well, obviously a card cannot have both a king and a queen, therefore, there would not be an intersection. Hope that makes sense!

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