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?
a) the probablity that the ball will both be white and have an even number is 0.
b) the probablity that the ball will both be white minus the probablity that the ball will have an even number on it is 0.2.

Probability ball: white - P(W)P(W);

Probability ball: even - P(E)P(E);

Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)−0P(WorE)=P(W)+P(E)−0. Not sufficient

(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 --> P(W)−P(E)=0.2P(W)−P(E)=0.2, multiple values are possible for P(W)P(W) and P(E)P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)P(WorE).

(1)+(2) P(W&E)=0 and P(W)−P(E)=0.2P(W)−P(E)=0.2 --> P(WorE)=2P(E)+0.2P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4P(E)=0.4 (10 even balls) then P(WorE)=1P(WorE)=1 BUT if P(E)=0.2P(E)=0.2 (5 even balls) then P(WorE)=0.6P(WorE)=0.6. Not sufficient.

E.

d

gimmee one of those gold crowns!

~coco

a orange at the grocery store idk what other product that could be