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What is the probability that event A occurs?

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What is the probability that event A occurs?

by gmattesttaker2 » Mon Apr 28, 2014 10:00 pm
Hello,

Can you please assist with this:

Events A and B are independent and have equal probabilities of occurring. What is the probability that event A occurs?

1.The probability that at least one of events A and B occurs is .84.
2.The probability that event B occurs and event A does not is .24.


OA: A


Thanks a lot,
Sri
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by vaibhav108 » Sun May 04, 2014 1:02 pm
Hello,

Can you please assist with this:

Events A and B are independent and have equal probabilities of occurring. What is the probability that event A occurs?

1.The probability that at least one of events A and B occurs is .84.
2.The probability that event B occurs and event A does not is .24.


[spoiler]OA: A[/spoiler]


Thanks a lot,
Sri
Given that, P(A) = P(B) and both are independent.

From 1 -
Atleast one of the events occur! That is either A will occur OR B will occur.
P(A) + P(B) = .84 ..... (OR represents Addition)
As, P(A) = P(B)
2P(A) = .84
P(A) = .42
SUFFICIENT

From 2 -
P(B) = .24 ....as A and B are independent.
[spoiler]NOT SUFFICIENT
[/spoiler]
[spoiler]ANSWER : A[/spoiler]
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by [email protected] » Sun May 04, 2014 8:12 pm
Hi Sri,

vaibhav108 made a slight error in his interpretation of this question, but his overall answer is correct.

We're told that events A and B are independent and have EQUAL PROBABILITY of occurring. We're asked for the probability that event A occurs.

Fact 1: The probability that AT LEAST one of the events occurs is .84

This means that A, or B or BOTH occurs = .84

We can restate this to mean that the probability that neither A, nor B nor BOTH occurs = .16

(Not A)(Not B) = .16

Since we're told that A and B have equal probabilities of occurring, they also have equal probabilities of NOT occurring. So....

(Not A) = (Not B)

Combining these 2 equations gives us the probability that A will not occur.....

(.4)(.4) = .16

The probability that A DOES NOT occur = .4
The probability that A DOES occur = .6
Fact 1 is SUFFICIENT

Fact 2: The probability that B occurs and A DOES NOT occur = .24

This tells us that....

(B)(Not A) = .24

From the prompt, we know that...

A = B
(Not A) = (Not B)
A + (Not A) = 1

In this scenario, the values are .4 and .6, but we don't know which is which (one is the probability the event occurs, the other is the probability that the event does NOT occur). Either one COULD be the answer.
Fact 2 is INSUFFICIENT.

Final Answer: A

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by vaibhav108 » Sun May 04, 2014 11:41 pm
Hi Rich,

Thanks for a great explaination!
vaibhav108 made a slight error in his interpretation of this question, but his overall answer is correct.
I think it was not slight error. It was a major one!

For 1 -
I did not consider the case that both A and B can occur. So P(A).P(B) needs to be considered.
Considering that case, I landed to a quadratic equation x^2-2x+.84=0, which is not that easy to solve. So, I will go with your method.

One quick question here...
We can restate this to mean that the probability that neither A, nor B nor BOTH occurs = .16
Shouldn't this be Probability of both (NOT A) AND (NOT B)? As we need to make sure that BOTH do not occur. Please correct me if I am wrong here...

Once again thank you for pointing out the flaw in my thought process!

-V
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by Matt@VeritasPrep » Mon May 12, 2014 12:24 pm
Here's a succinct approach:

Let's say the probability of A occurring is x. Then the probability of A NOT occurring is (1 - x).

Since A and B have identical probabilities, B's probabilities are also x and (1 - x), respectively.

S1 tells us that

1 - (Probability of Neither A Nor B) = .84

or

1 - (1 - x)² = .84

or

(1 - x)² = .16

or

(1 - x) = .4

Since (1 - x) is the probability of B's NOT occurring, we know the probability of B's occurring is .6, or 60%, and S1 is SUFFICIENT!

S2 tells us that

x * (1 - x) = .24

or

x - x² = .24

or

x² - x + .24 = 0

or

(x - .6)(x - .4) = 0

This gives us TWO distinct, valid solutions for x, so it's INSUFFICIENT.
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by GMATinsight » Tue May 13, 2014 7:24 am
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