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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]  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]  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]  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 [ 4.17 KiB | Viewed 23756 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]  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]  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]  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]  18 Dec 2018, 08:43
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]  18 Dec 2018, 08:54
Expert's post
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
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: In a probability experiment, G and H are independent events [#permalink]  26 Dec 2018, 13:22
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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]  27 Jan 2019, 05:15
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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.

Cheers,
Brent
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Re: In a probability experiment, G and H are independent events [#permalink]  09 Sep 2019, 18:55
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

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.

Practice Questions
Question: 5
Page: 119

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.
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Re: In a probability experiment, G and H are independent events [#permalink]  14 Jun 2020, 11:48
GreenlightTestPrep wrote:
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.

Cheers,
Brent

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.
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Re: In a probability experiment, G and H are independent events [#permalink]  14 Jun 2020, 11:54
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Asmakan wrote:
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.

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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Re: In a probability experiment, G and H are independent events [#permalink]  15 Jun 2020, 01:40
GreenlightTestPrep wrote:
Asmakan wrote:
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.

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?

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
Re: In a probability experiment, G and H are independent events   [#permalink] 15 Jun 2020, 01:40
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