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

Not sufficient information

Step-by-step explanation:

Probability of white ball: P(W)

Probability of even number: P(E)

Probability of white ball or even number: P(W∨E) is asked

Considering case (1):

Probability of white ball and even number: P(W∧E) = 0

And we know that,

P(W∨E) = P(W) + P(E) - P(W∧E) = P(W) + P(E) - 0 = P(W) + P(E)

It is not sufficient to calculate the desired probability.

Considering case (2):

P(W) - P(E) = 0.2

Here, it is possible for P(W) and P(E) to get multiply values.

So P(W∨E) cannot be determined.

Considering cases (1) & (2):

P(W∧E) = 0 and P(W) - P(E) = 0.2

So, P(W∨E) = P(W) + P(E) = P(E) + 0.2 + P(E) = 2P(E) + 0.2

Again multiple answers are possible.

Such as, for P(E) = 0.4 (10 even balls) ⇒ P(W∨E) = 1

but for P(E) = 0.2 (5 even balls) ⇒ P(W∨E) = 0.6

So, the information are not sufficient.

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step-by-step explanation: