Explanation

By the rules 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 \(s(1 - r)\). So Quantity A, the probability that either G will occur or H will occur, but not both, is \(r(1 - s) + s(1 - r) = r + s - 2r s\), which is less than Quantity B, \(r + s - r s\). Thus the correct answer is Choice B.
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I struggled a bit with understanding the mathematics in this question, but I think I found a more intuitive way to explain the solution.



Let the RED area equal r, or P(G), and let the BLUE area equal s, or P(H).

The PURPLE area, therefore, indicates the probability that G and H both occur, or P(G intersection H).

Since we're trying to find "The probability that either G will occur or H will occur, but not both," then we are trying to find the RED area and the BLUE area, but NOT the PURPLE area. To write this in probability terms, we would write P(G union H) - P (G intersection H) = P(G) - P (G intersection H) + P(H) - P (G intersection H)

The RED area is P(G)-P(G intersection H), or r-rs. The BLUE area is P(H)-P(G intersection H), or s-sr. You then simply add those too formulas together, resulting in r - rs + s - rs, which simplifies to r + s - 2rs.

May I know why the formula P(G or H)= P(G)+P(H)-P(G&H) does not apply here. Thanks for the clarification.

Here is one query(not related to above question), if the quantities are not mutually exclusive, then which formula should we use?

Formula for mutually exclusive P(A or B)= P(A) + P(B)

Answer: B
G and H are independent events.
The probability that G will occur is r: P(G) = r > 0
The probability that H will occur is s: P(H) = s > 0
A: The probability that either G will occur or H will occur, but not both:
P(G delta H) = P(G and not H) + P(H and not G) = P(G)*P(H’) + P(H)*P(G’) = P(G)*(1- P(H)) + P(H)*(1- P(G)) = P(G) - P(G)*P(H) + P(H) -P(G)*P(H) =
P(G) + P(H) - 2P(H intersection G) = [since G and G are independent, P(H intersection G) equals probability of multiplication of them: P(G)*P(H)
= P(G) + P(H) - 2P(G)*P(H) = r + s - 2r*s

B= r + s - r*s
So B is bigger than A.
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