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Probability / Gmatprep CAT

by koshls » Tue Aug 28, 2012 3:24 am
Each of the 25 balls in a certain box is either red, blue or white and has a number 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

1) The probability that the ball will both be white and have an even number painted on it is 0.

2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2

[spoiler]OA: E[/spoiler]

Pls help explain. Thanks
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by Anurag@Gurome » Tue Aug 28, 2012 3:30 am
koshls wrote:Each of the 25 balls in a certain box is either red, blue or white and has a number 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

1) The probability that the ball will both be white and have an even number painted on it is 0.

2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2

[spoiler]OA: E[/spoiler]

Pls help explain. Thanks
The required probability = Probability that the selected ball is white + Probability that the selected ball that have an even number painted on it = P(white) + P(even) - P(white & even)

We are subtracting P(white & even) as the balls that are white as well as have an even number painted on it has been calculated twice.

We need to know these two probabilities.

Statement 1: P(white & even) = 0
Hence, these two events are exclusive, i.e. none of the white balls have an even number on it. Therefore, The required probability = P(white) + P(even)

This statement helps to reduce the number of unknowns but we don't have enough information yet to find the required probability.

Not Sufficient

Statement 2: P(white) - P(even) = 0.2
Not enough information to determine the required probability.

Not Sufficient

1 & 2 Together: We can't determine [P(white) + P(even)] just from knowing that [P(white) - P(even)] = 0.2

Not Sufficient

The correct answer is E.
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by vk_vinayak » Tue Aug 28, 2012 3:41 am
P(white OR even) = P(white) + P(even) - P(white & even)
P(x OR y) = P(x) + p(y) - P(x AND y)

Can we use this formula whenever we see P(x OR y) ?
- VK

I will (Learn. Recognize. Apply)
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by Brent@GMATPrepNow » Tue Aug 28, 2012 6:50 am
koshls wrote:Each of the 25 balls in a certain box is either red, blue or white and has a number 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

1) The probability that the ball will both be white and have an even number painted on it is 0.

2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2

[spoiler]OA: E[/spoiler]
Target question: What is the value of P(white or even)?

To solve this, we need P(A or B) = P(A) + P(B) - P(A & B)
So, P(white or even) = P(white) + P(even) - P(white & even)

Statement 1: P(white & even) = 0
We can add this to our probability equation to get: P(white or even) = P(white) + P(even) - 0
Still need P(white) and we need P(even)
INSUFFICIENT

Statement 2: P(white) - P(even)= 0.2
We have no idea about the sum of P(white) and P(even), and we don't know the value of P(white & even)
IINSUFFICIENT

(1)&(2) Given P(white) - P(even)= 0.2 does not tell us the individual values of P(white) and P(even), and it doesn't tell us the value of P(white) + P(even).

So, since we can't determine the value of P(white) + P(even) - P(white & even), the statements combined are not sufficient.

Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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