SOLUTION: Suppose D, E, and F are independent events with P(D) = 0.21 , P(E) = 0.48, and P(F) = 0.18. What is the probability that at least one of the three events occurs?

Question 413498: Suppose D, E, and F are independent events with P(D) = 0.21 , P(E) = 0.48, and P(F) = 0.18. What is the probability that at least one of the three events occurs?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
The probability that at least 1 of the events occur is equal to 1 minus the probability that none of the events occur.

p of d = .21
p of e = .48
p of f = .18

p of not d = 1 - .21 = .79
p of not e = 1 - .48 = .52
p of not f = 1 - .18 = .82

p of none of the events occurring is equal to .79 * .52 * .82 = .336856

p of at least one of the event occurring would be 1 -.336856 = .663144

to see if this is good, just take the possibility of 1, 2, or 3 of the events occurring and add them up.

it's a lot more labor intensive to do it this way, but it is instructive.

because each of the probabilities are different, you can't take any shortcuts.

you have to find:

p of only first event occurring.
p of only second event occurring.
p of only third event occurring.

p of only first and second event occurring.
p of only first and third event occurring.
p of only second and third event occurring.
p of all 3 events occurring.

then you have to add them up.

the number required are shown below:
.089544 = p of only first event occurring.
.310944 = p of only second event occurring.
.073944 = p of only third event occurring.

.082656 = p of only first and second event occurring.
.019656 = p of only first and third event occurring.
.068256 = p of only second and third event occurring.
.018144 p of all 3 events occurring.
sum of all these probabilities is equal to .663144

That's the same we got the easy way by taking 1 - p that none of the events occur.

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