Explanation:
By the rule of probability, you can conclude that the probability that event H will not occur is 1-s. Also, 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(2-r). So Quantity A, the probability that either G will occur or H will occur, but not both, is r(1-s) + s(1-r) = r+s - 2rs, which is less than Quantity B, r+s -rs. Thus the correct answer is Choice B.


What confuses me is why you can't just use the equations:
P(H or G) = P(H)+ P(G) - P(H and G)
P(H and G) = P(H)P(G) (because they are independent)

so P(H or G) = p(H) + P(G) - P(G)p(H) ---> P (H or G) = s +r - rs

Why doesn't it work in this case? Another problem (4.44 - data analysis pg 318) showed a similar case where this formula was used:
Consider an experiment with events A, B, and C for which P(A) = 0.23, p(B) = 0.4, and p(C) = 0.85. Suppose that events A and B are mutually exclusive and events B and C are independent. What is P (B or C)?
Since B and C are independent, P(B and C) = P(B)P(C).
So P(B or C) = P(B)+P(C) - P(B)P(C).
- P(B or C) = 0.40 + 0.85 - (0.4)(0.85) = 1.25 - 0.34 = 0.91

Couldn't you think of G as B with r= 0.4 and H as C with s=0.85. If that were the case and you plugged these values into the official answer for problem 5, you would get the following.
r + s - 2rs
0.4 + 0.85 - 2(0.4)(0.85) =0.57 which isn't the answer 0.91.


I'm sorry for the confusing long question. I just don't understand how it would work in one instance but not in another and when I would use r + s - 2rs vs r + s - rs .