Question 413498
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.