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).

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)

Your mistake is the assumption that the numbers are written equally.

"Each ball has a number from 1 to 10 painted on it" does NOT tell us that each number is used an equal number of times (with 25 balls, that would actually be impossible) or what the distribution is.

In fact, according to the original info (and even the statements), it's possible that EVERY ball has a "3" on it.