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In a probability experiment, G and H are independent events

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In a probability experiment, G and H are independent events [#permalink] New post 15 Jun 2016, 15:53
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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] New post 15 Jun 2016, 15:56
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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] New post 17 Jun 2016, 20:09
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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.
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Re: In a probability experiment, G and H are independent events [#permalink] New post 22 Apr 2017, 13:47
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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
Probability.png [ 4.17 KiB | Viewed 7242 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.
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Re: In a probability experiment, G and H are independent events [#permalink] New post 03 Dec 2017, 08:31
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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.
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Re: In a probability experiment, G and H are independent events [#permalink] New post 23 Aug 2018, 21:19
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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.
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Re: In a probability experiment, G and H are independent events [#permalink] New post 26 Aug 2018, 01:58
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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] 26 Aug 2018, 01:58
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