[email protected] 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 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?

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

II] 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 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.