This isn't actually a homework question, I'm reviewing for a midterm I have coming up this week and came across this question in some of the practice exercises that were provided.
Let A,B,C be three events. Suppose that A and B are independent, B and C are mutually exclusive, and that A and C are also independent. Given that P(A)=0.3, P(B)=0.4, P(C)=0.3 find the probability that at least one of the three events occur.
For independent events / non-mutually exclusive P(A union B) = P(A) + P(B) - P(A intersect B)
P(A intersect B) = P(A)P(B)
for mutually exclusive events P(A union B) = P(A) + P(B)
The Attempt at a Solution
I know the correct answer is 0.79 (solutions provided) but I'm having trouble understanding how to approach the combination of independent / mutually exclusive events.
Is it easier to count the complement? ie:the event that none of the 3 occurs and subtract it from 1?
If I sum the probabilities of A and B and the probability of A AND C I get 0.79.
ie: 0.3 + 0.4 + 0.3(0.4), but is this the correct method?
Thank you in advance if you can help me understand this!