OG quantitative guide pr. 5, p. 109

[irs031]

Problem 5:
In a probability experiment, G and H are independent events. The probability that G will occur is r, and the probability that H will occur is s, where both r and s are greater than 0.

Q A: The probability that either G will occur or H will occur, but not both
Q B: r+s - rs
I chose C and I was wrong.
The official solution: The fact that G and H are independent events implies that G and "not H" are independent events. Therefore the probability that G will occur and H will not occur is r(1-s). Similarly the probability that H will occur and G will not occur is s(1-r).
r(1-s) + s(1-r) = r+s - 2rs, thus correct answer is B; B is greater than A.

I can not really understand what necessitates the multiplication of the prob that G will occur and H not occur?

What I also do not understand is how this problem is different from pr. 3 on p. 132 in guide 5 "Word problems":

A fair die is rolled and a fair coin is flipped. What is the probability that either the die will come up 2 or 3, OR the coin will land up heads up?
Solution: probability that die will come up 2 or 3 is 1/3; probability that the coin will land heads up is 1/2; probability that both events will happen is (1/3)(1/2) = 1/6

Therefore: 1/3+1/2 - 1/6 = 2/3

In this case why we do not need to multiply prob of die coming up 2 or 3 to prob of coin NOT coming heads up (or vice versa)?

[tommywallach]

Hey IRS,

Maybe the probabilistic wording will help. To find Quantity A, we're happy if:

G occurs AND H doesn't OR H occurs AND G doesn't.

Remember, anytime you have the word "AND," you multiply, and any time you have the word "OR" you add. That's where this math comes from.

In the other example you give, you could solve it the same way:

Die comes up 2 or 3 AND coin lands heads down OR Die comes up 1,4,5,6 AND the coin lands heads up OR die comes up 2 and 3 AND coin lands heads up (because that isn't removed as a possibility either either/or language):

1/3 * 1/2 + 2/3 * 1/2 + 1/3 * 1/2 =
1/6 + 2/6 + 1/6 = 4/6 = 2/3

The difference is that the possibility of both successes occurring is still allowed. The 1/6 is being subtracted because we've double counted the possibility when we add 1/3 + 1/2.

-t

[irs031]

Thanks to your thorough explanation and me rereading the chapter on Probability in the Word problems guide, now I believe I understand the problem.
The official guide problem is asking for the probability of event r occurring OR event s occurring, BUT NOT BOTH. For that reason the prob that event r AND event s occur simultaneously is subtracted twice (r+s - 2rs). Similarly if the other problem I gave as an example on p. 132 of the "Word problems" guide was asking "What is the probability that either the die will come up 2 or 3, OR the coin will land heads up, BUT NOT BOTH" then the solution would have been (1/3)+(1/2) - (2)(1/6) = 1/2
OR
(1/3)(1/2) + (2/3)(1/2) = 1/6+2/6 = 1/2

If the problem is asking for the probability of event A OR event B occurring, then that means prob that A occurs, B doesn't, B occurs A doesn't or A and B both occur.
If the problem is asking for the probability of event A OR B occurring, but NOT BOTH, than this is prob of A occurs, B doesn't or A occurs, B doesn't only.
Small difference in wording, but crucial to solve the problem.
Tricky problem.

Thanks Tommy.

[tommywallach]

Exactly right, IRS! And glad to help!

-t