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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # In a probability experiment, G and H are independent events  Question banks Downloads My Bookmarks Reviews Important topics
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In a probability experiment, G and H are independent events [#permalink]
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Question Stats: 37% (00:43) correct 63% (00:44) wrong based on 100 sessions
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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Question: 5
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[Reveal] Spoiler: OA

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Re: In a probability experiment, G and H are independent events [#permalink]
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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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Re: In a probability experiment, G and H are independent events [#permalink]
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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. Intern Joined: 22 Apr 2017
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Re: In a probability experiment, G and H are independent events [#permalink]
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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.

Attachment: Probability.png [ 4.17 KiB | Viewed 9993 times ]

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. Intern Joined: 03 Dec 2017
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Re: In a probability experiment, G and H are independent events [#permalink]
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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. Intern Joined: 12 Aug 2018
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Re: In a probability experiment, G and H are independent events [#permalink]
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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. GRE Prep Club Legend  Joined: 07 Jun 2014
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Re: In a probability experiment, G and H are independent events [#permalink]
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Runnyboy44 wrote:
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.

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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Re: In a probability experiment, G and H are independent events [#permalink]
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)
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Re: In a probability experiment, G and H are independent events [#permalink]
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Sawant91 wrote:
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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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
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Re: In a probability experiment, G and H are independent events [#permalink]
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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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Re: In a probability experiment, G and H are independent events [#permalink]
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sandy wrote:
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

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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