The original question was:
"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."
Now I want to show that the 2 statements both together are insufficient. So let's suppose they are both true.
If the ball is white there are two cases: either the ball is white with an odd number painted on it or the ball is white with an even number painted on it (but we know that the second case is impossible due to the first statmenet.
If the ball has an even number painted on it there are two cases: either the ball is white or it is not, but we know that the first case is impossible due to the first statement. So what the second statement is very saying is "The probability that the ball will be white and will not have an odd number on it minus the probability that the ball will not be white and will have an even number painted on it is 0.2". So Let's look at the table below. I called A the prob. "not white, even number", I called B the probabililty "not white odd number". As you can see the only condition we have is that A+A+0.2+B=1 with 0<=A,B<=1 but there are different several ways to do that: A=0.1, B=0.6 rather than A=0.2, B=0.4 . So they are insufficient.
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