Probability Question

Are you aware of P(A U B) = P(A) + P(B) - P(A n B)?

I don't understand the subtraction of the probability both occur? If the question had asked under 1: the probability either event occurs I would have subtracted the possibility of both, but I don't understand how in this scenario its been counted twice?

You know the probability of A is 0.3 and B is 0.2, and that they are independent. You have a number of possible outcomes:

1. A only
2. B only
3. A & B
4. Neither A nor B

P(1)=0.3 * (1-0.2) =0.24
P(2)=(1-0.3) * 0.2 =0.14
P(3)=0.3 * 0,2 =0.06
P(4)=(1-0.3)*(1-0.2)=0.56

Adding them, you will see that we get 1.0, as we've covered all the possibilities. If you add the 0.3 and 0.2, you get 0.5, but the probability is actually 0.24+0.14+0.06=0.44, as the 0.5 counts the probability of both occurring twice - once in the 0.3 and once in the 0.2. Think about a 0.2 and 0.3 overlapped by 0.06.

For example: in a problem where a die is rolled and a coin is flipped, if you are asked for probability that either the die will come up 2 or the coin will land on heads you would add the probability of either event and then minus the probability of both (but in this question it specifically asks you to exclude both and not include it like the one above.)

Thank you very much and apologies this is at such a basic level. I think my confusion has stemmed from me conceptually thinking that the individual probabilities added includes the possibility of them both occurring?

Isn't that what you need to do here, i.e. 0.2 + 0.3 - 0.2*0.3 = 0.44?

You're essentially overlapping 0.2 and 0.3 by 0.06 in the probability range, so you need to remove 0.06 to get the correct probability or either / both.

In the coin/die problem it specifically asks you to exclude the possibility of both, whereas the original problem asks you to include this.

I never did like probability, so it's good to have to think about it a bit.


Where did it say that? I'm not seeing it.

In the rain and test problem: chance of rain AND/OR test: 3/10 + 2/10 = 5/10. But apparently this includes chance of both counted twice (im not sure how), so remove one of them 3/10 x 2/10 = 0.06. 0.5 - 0.06 = 0.44.

In the coin/die problem: chance of a heads OR a 2. 1/2 + 1/6 = 4/6. But need to remove chance of both (4/6 - (1/2 x 1/6)). But from reasoning above, if you add both the probabilities, this includes the chance of both x 2, so wouldnt you need to remove 2 x the chance of both?

The probabilities overlap. Think of two coin tosses - what's the probability of a heads on the first throw and / or on the second one? If you add, then you'd get 1.0 - clearly wrong. The correct answer is 0.75, as the 0.5 and 0.5 overlap by 0.25. We can only count that 0.25 once, getting (0.5-0.25) + 0.25 + (0.5-0.25) = 0.75.


You've removed (once) the probability of both in each of those examples.

You only remove the chance of both once, because you still want it to be included (once) in the final answer.