8% chance A will happen.
12% chance B will happen.
30% chance C will happen.

For the sake of this thread assume A, B and C are completely independent of each other.

Is my overall chance of A, B or C happening 50% or one-in-two?

The reason I ask is last night I was watching something that was spouting all the statistical chances of all sorts of things that would kill you. From shark attacks and lightning strikes to auto accidents and slipping in the bathtub. When they were all done it felt as if you added them all together almost no one should make it to old age yet that clearly isn’t the case. What am I missing.

Well, you certainly can’t add them together. If you could, you couldn’t walk out the door:

8% chance A will happen.
12% chance B will happen.
30% chance C will happen.
20% chance D will happen.
10% chance E will happen.
4% chance F will happen.
26% chance G will happen.

An easier way of determining if any one or more of A, B, and C will happen is to calculate the odds that NONE of them at all will happen. From your example:

So the probability that A will NOT happen is 92%, the probability that B will NOT happen is 88%, and the probability that C will NOT happen is 70%. The probability that NONE of them will happen (provided these are independant) is 92% X 88% X 70% = 56.7%

From there it’s an easy jump: If the probability that neither A nor B nor C will happen is 56.7%, then the probability that AT LEAST ONE of them will happen is (1 - .567) = 43.3%.

I think the answer depends on how we take the term “completely” independent.

Two or more events are either independent or dependent. What then does the word “completely” do?

Without other clues as to the meaning, I would take the OP to actually mean mutually exclusive. That is, there is no chance any two of the events can occur. Therefore I would simply add the probabilities to get the answer that one will.

<< there is no chance any two of the events can occur. Therefore I would simply add the probabilities to get the answer that one will. >>

This is the same as the example offered by cabbage and by zut, except that the probability of both A and B is zero.

Summarizing what everyone has said:

Example 1:
Probability of rolling 1 on one roll of a 1-sided die = 1/6
Probability of rolling 2 on one roll of a 1-sided die = 1/6
Probability of rolling 1 or 2 on one roll of a 1 sided die = 1/6 + 1/6 = 1/3 (aahala’s method)
Probability of rolling BOTH 1 AND 2 on one roll = 0
Probability via cabbage’s method = 1/6 + 1/6 - 0 = 1/3

Example 2:
Probability of getting a 1 on the first roll of a die: 1/6
Probability of getting a 1 on the second roll of a die: 1/6
However, the probability of rolling at least one 1 on two rolls of a 1-sided die is NOT 1/6 + 1/6. (Think: if you roll the die 6 times, you are NOT assured of rolling a 1)

In this case, the two events are NOT “completely” independent. It is possible to have BOTH a 1 on the first roll AND a 1 on the second roll. Hence, use cabbage’s formula, although it’s easier to use zut’s approach: Probability of not getting 1 on the first roll = 5/6
Probability of not getting 1 on the second roll = 5/6.
Probability of not getting 1 on either roll = 25/36, so probability of at least one 1 = 1 -25/36 = 11/36, slightly less than 1/3.

That “slightly less” is why it’s possible to rolll the die six times and not hit a 1 in any roll.

General rule way oversimplified: if the events are truly “completely independent”, then add probabilities to get the odds for EITHER and multiply probabilities to get the odds for BOTH.

To add to what’s been said, the only way that the calcluation in your OP is correct is if the events A, B, and C are pairwise mutually exclusive. A and B are independent iff P(A and B) = P(A)P(B). A and B are mutually exclusive iff P(A and B) = 0.