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probability

by sunilrawat » Fri Sep 23, 2011 9:18 pm
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is drawn at random, what is the probability that the ball 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 is 0.
2. The probability that the ball will be white minus the ball will have an even number painted on it is 0.2

OA E
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by cans » Fri Sep 23, 2011 9:44 pm
even numbers: 5 (2,4,6,8,10)

A) both white and even=0. insufficient
B) p(white)-p(e) = 0.2
but we need to find p(w) + p(e) - p(ew)
insufficient
IMO E
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by Nidhi4mba » Sat Sep 24, 2011 1:09 am
I agree with Cans. Answer should be E. What's the OA??
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by apex231 » Sat Sep 24, 2011 6:36 am
I think together A and B are sufficient.

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

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

n(W) = number of white balls
n(E) = number of balls with even nbr printed = 5

From stmt 1 we know that no white ball has an even nbr printed on it.

From stmt 2, P(W) - P(E) = .2

stmt 1 and 2 together -

n(W)/25 - n(E)/25 = 0.2 -> n(W) - n(E) = 5 -> n(W) - 5 = 5 -> n(W) = 10

probability that the ball will be either white or have an even number printed on it =
(n(W) + n(E))/25 = (10 + 5)/ 25 = 15/25 = 3/5
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by Brent@GMATPrepNow » Sat Sep 24, 2011 7:22 am
sunilrawat wrote:Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is drawn at random, what is the probability that the ball 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 is 0.
2. The probability that the ball will be white minus the ball will have an even number painted on it is 0.2

OA E
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)

Our goal is to find the value of P(white or even)

(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)
INSUFF

(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)
INSUFF

(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

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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