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I have a set of binary strings, length $8$ bits, so I have that the probability that these strings have two bits equal to $11$ in the least significant positions is $1/4$ (event $A_1$)

If I want to know the probability that these strings have $2$ bits equal to $11$ in the most significant position is $1/4$ also (event $A_2$)

I would like to know why it appears in the book that I am reading that:

$$\Pr[A_1 \cup A_2] < \Pr[A_1] + \Pr[A_2]$$

from what I got the union of those probabilities is $1/2$ which is equal to the sum of their individual probabilities.

Am I missing something?

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Hint: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. What does the first axiom of probability theory say about the probability of any event? Is it possible for your chosen string to have $11$ in both the most significant two bits and the two least significant bits? If so, what could you say about $P(A_1 \cap A_2)$? – Dilip Sarwate Jan 19 '13 at 16:31
I could say that the intersection is not equal to null, so in that case the prob of the union will be less than the sum of the individual probabilites, am I right? – Manolo Jan 19 '13 at 16:41
It might be better to say that the intersection is not the empty set, but yes, you have found the correct reason why $P(A_1\cup A_2)$ is strictly smaller than $P(A_1)+P(A_2)$. – Dilip Sarwate Jan 19 '13 at 19:46
And now that you have found it, you can write it up as an answer to your question (this is encouraged on this website), and then after some time has passed you can accept your own answer. – Gerry Myerson Jan 20 '13 at 11:54

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