Stuart Kovinsky wrote:dextar wrote:Each of the 25 balls in a certain box is either red, blue or white. Each ball has a number from 1 to 10 painted on it. If one ball is selected at random, the probability that it is either white or has an even number on it is?
1)Probability of selecting a white ball having an even number is 0
2)Probability that the ball be whiteminus theProbability that the ball be even=0.2
Pls see whether we are getting the answer to this problem or not.
We know that probability = # of desired outcomes / total # of possibilities.
In this case, we know that the total # of possibilities = 25, so we need to know the # of balls that are white and/or have an even number on them.
(1) there are no white even balls - doesn't tell us how many there ARE, so insufficient.
(2) If there's an extra .2 chance of getting a white over an even, and we know that there are 25 balls total, we know that white - even = 5 (20% of 25). However, we could have 15 white and 10 even or 8 white and 3 even or lots of other combos: insufficient.
Combined: we could still have 15/10 or 8/3, so we still have no clue how many desired outcomes exist: choose (e).
THanks for ur explanation but I have another approach here. Pls tell me where I'm wrong.
I think I'm not interpreting the question stem wrong when it says
"Probability that the ball be whiteminus theProbability that the ball be even=0.2".
What I can infer from this is P(Ball is white)-P(even ball)=0.2
Since we know that the numbers are from 1 to 10 so P(even ball)=5/10=0.5
So P(Ball is white)=0.7
From 1)
P(Ball is White and Ball is even)=0
So, P(Ball is white
or Ball is even)=P(Ball is white) + P(Ball is even)-P(Ball is White
and Ball is even)
Putting these values I'm geting P(Ball is white or Ball is even)=0.7 +0.5
=1.2
(which is not possible)
