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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 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 zero. (2) The probability that the ball will be white minus the probability that the ball have an even number painted on it is 0.2.
Pw - the probability that the ball will be white Pe - the probability that the ball have an even number painted on it Pwe - the probability that the ball will both be white and have an even number painted on it Pw|e - the probability that the ball selected will either be white or have an even number painted on it
both Pw=0.3, Pe=0.1 and Pw=0.5, Pe=0.3 satisfy two conditions but
Pw=0.3, Pe=0.1: Pw|e = Pw + Pe = 0.4 Pw=0.5, Pe=0.3: Pw|e = Pw + Pe = 0.8
As per question we need white OR even (we DON'T want white AND even - we have to be able to strip out those that fall into both categories). To calculate, can either do: a) probability of "white and odd" + probability of "even and not white" b) probability of white + probability of even - probability of white & even
Statetment 1: Tells us there aren't any that are both white and even. This doesn't tell us how many are white or how many are even. Insufficient. Eliminate A and D.
Statement 2: Tells us that Pwhite - Peven = 0.2. So, Pwhite could be 0.4 which would make Peven 0.2. Or Pwhite could be 0.3 which would make Peven 0.1. And (by itself) it doesn't tell me Prob of even & white, which I'd need to subtract, so... insufficient in many ways. Eliminate B.
Combining both statement we know that Peven+white = 0. BUT, we still have multiple possibilities for Pwhite and Peven (see above). 0.4+0.2-0=0.6. 0.3+0.1-0=0.4. ?? Still insufficient.