It is valid:

Probability that either event occurs = \(P(G)+P(H)-P(G&H)\)

Probability that either event occurs but not both = \(P(G)+P(H)-P(G&H)-P(G&H)= r+s-2rs\). This is quantity A
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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.

Let A and B are two events,

P(A U B) - Probability that either A will occur OR B will occur OR both will occur

P(A ∩ B) - Probability that A and B both will occur

P(A U B) = P(A) + P(B) - P(A ∩ B)

If A and B are independent events, which means they are not related to each other and occur independent of each other,
then P(A ∩ B) = P(A)*P(B)

P(G) = r
P(H) = s

Quantity A -
= P(G U H) - P(G ∩ H) = [P(G) + P(H) - P(G ∩ H)] - P(G ∩ H) = [r + s - rs] - rs = r + s - 2rs < r + s - rs

Hence, Answer is B

A:
P(G union H) - P (G intersection H) = P(G) + P(H) - P (G intersection H) - P (G intersection H) = r + s -rs -rs = r + s -2rs

B:
r + s -rs

Clearly B is greater, hence the answer.

For the official guide answer to be correct you have to assume G & H are not mutually exclusive events (events that can't both happen at the same time. If they were then R*S would equal 0 and A and B would be equal.

You have to be able to infer that since the probability of G happening doesn't effect the probability of H happening AND that r and s are greater than zero that G & H must not be mutually exclusive.

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)

HI,

Standard formula is P(A or B)= P(A) + P(B)-P(A and B)
Mutually exclusive means that the two events cannot happen at the same time, so intersection is 0, that is P(A ∩ B) = 0, so we are left with P(A or B)= P(A) + P(B)-P(A and B) = P(A) + P(B)-0 = P(A) + P(B)
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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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Since G and H are independent events, we know that P(G and H) = rs
Also, since we're told that r and s are each greater than 0, we know that rs > 0
In other words, P(G and H) > 0

Now recognize that P(G or H) = P(G) + P(H) - P(G and H)
= r + s - rs
Aha! r + s - rs = P(G or H)
IMPORTANT: P(G or H) = P(G occurs, or H occurs or BOTH G and H occur)

So, we have:
QUANTITY A: P(G occurs, or H occurs not NOT BOTH G and H occur)
QUANTITY B: P(G occurs, or H occurs or BOTH G and H occur)

Since P(G and H) > 0, we can conclude that Quantity B is greater.

Answer: B

Cheers,
Brent
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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.

I have a question:
Based on what you put,
Example :
A:
r= 1.5, s= 1.5, Both will happen =0
P( R or S )= 1.5+1.5 - 0 = 3

B;
Based on the previous, the value is 3.
May you please clarify your explanation?

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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Sorry, I was confused with another question.
Yes, I agree with u, probability can't be greater than 1.
In Q A, It is mentioned that the 2 probability don't happen with each other, " but not both ". So, the probability here is 0