probability

[sunilrawat]

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

[cans]

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

[Nidhi4mba]

I agree with Cans. Answer should be E. What's the OA??

[apex231]

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

[Brent@GMATPrepNow]

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

Cheers,
Brent