Each of the 25 balls in a certain box is either red, blue,

[BTGmoderatorLU]

Source: GMAT Prep

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 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.

The OA is E.

[GMATGuruNY]

Source: GMAT Prep

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 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.

Let:
W = the total number of white marbles.
E = the total number of even marbles.
WE = the total number of white even marbles.

Statement 1:
Thus, WE = 0.
No way to determine the value of W or E.
INSUFFICIENT.

Statement 2:
Implication:
W-E = (0.2)(25) = 5.

Case 1: W=10, E=5, WE=0
Here, P(W or E) = (10+5)/25 = 15/25 = 3/5.

Case 2: W=11, E=6, WE=0
Here, P(W or E) = (11+6)/25 = 17/25.

Since P(W or E) can be different values, INSUFFICIENT.

Statements combined:
Cases 1 and 2 satisfy both statements.
Since P(W or E) can be different values, INSUFFICIENT.

The correct answer is E.

[Brent@GMATPrepNow]

Source: GMAT Prep

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 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.

The OA is E.

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)
INSUFFICIENT

Statements 1 and 2 combined:
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