Stiv wrote:
Bunuel 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?
Probability ball: white - \(P(W)\);
Probability ball: even - \(P(E)\);
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) --> \(P(WorE)=P(W)+P(E)-0\). 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 --> \(P(W)-P(E)=0.2\), multiple values are possible for \(P(W)\) and \(P(E)\) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine \(P(WorE)\).
(1)+(2) \(P(W&E)=0\) and \(P(W)-P(E)=0.2\) --> \(P(WorE)=2P(E)+0.2\) --> multiple answers are possible, for instance: if \(P(E)=0.4\) (10 even balls) then \(P(WorE)=1\) BUT if \(P(E)=0.2\) (5 even balls) then \(P(WorE)=0.6\). Not sufficient.
Answer: E.
Hope it's clear.
You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2)
Are these values only possible? If they are please explain why.
Why they cannot be 0.3 and 0.1 or 0.5 and 0.3?
solution:
White balls = w
Red = R
Blue = b
Total ball = 25
Sum total of even numbered ball = E
We have to evaluate = w/25 + (E)/25
From st(1) , we only know there is no white ball which contains even number. Even we still don’t know about red and blue balls have how many even numbered balls in them. So all are in mystery and doubt.
From st(2), w/25 – E/25 = 0.2 , again all unknown.
From both statement, we can assume several different things, I mean double case.
So both are insufficient. Answer is (E)