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.

Quantity A	Quantity B
The probability that either G will occur or H will occur, but not both	r + s - rs

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.




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Sandy
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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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Follow your heart

A quick way of doing this is imagining a Venn diagram. Since these events are independent, we know that they must have some overlap (if they were mutually exclusive, there would be no overlap in our Venn diagram. By adding both A and B. We count the overlapping region twice. Thus RHS is how we could account for the probability of either event happening, but the LHS would no include that region.

So clearly B is larger than A.

There are a few issues with your question above.

First of all, the probability of an event is always less than or equal to one.
So, r and s cannot equal 1.5
Also, where does the zero come from in 1.5+1.5 - 0 = 3?
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Brent Hanneson – Creator of greenlighttestprep.com
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