$\begingroup$P(A) = 0.7, P(B) = 0.2 assuming independent events
what is the probability of not A or not B?
I thought I could substitute P(A') = 1-P(A) and P(B') = 1-P(B) so:
P(A' or B') = 1-P(A) + 1-P(B) = 0.3 + 0.8 = 1.1
but 1.1 is greater than 1 so that can't be the correct answer can someone help me with what I did wrong?
$\endgroup$51 Answer
$\begingroup$You say that A and B are independent but you do not say wheter or not they are mutually exclusive!
In general, P(A or B)= P(A)+ P(B)- P(A and B). If A and B are disjoint so that P(A and B)= 0 then P(A or B)= P(A)+ P(B).
Here, you say that P(A)= 0.7 and p(B)= 0.2. Those add to 0, so it is quite possible (but not certain) that they are mutually exclusive. But P(A')= 0.3 and P(B')= 0.8. Those add, as you saw, to 1.1> 0 so even if A and B are mutually exclusive, A' and B' cannot be!
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