there are 25 balls painted red white or blue and each ball has a number from 1- 10 painted on it. If one ball is selected at random what is the probability that the ball will be white or have an even number painted on it?
1)the probability that the ball will be both white and have an even number is zero
2)the probability that the ball will be white minus the probability that the ball will be even is 0.2
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The question is asking you, the probability of picking a white or an even number. This means Probability of(white+even).
With no information, the number of white balls can be anything from 1-23 (assuming one is red and one blue).
The number of even can be anything from 0-25.
To calculate number of total balls which are either white or even, you should know, number of balls only white, number of balls only even and number of balls which are both even and white. The sum of these divided by 25 will give you the probability.
i) None of the white balls has an even number.
It is possible that there are 15 white balls and the rest 10 have even number, or it is possible that none of the 25 balls has even number. The answer can be 25/25=1, or 15/25, etc. INSUFFICIENT.
ii) Difference of probability of a white ball and an even numbered ball is 0.2 or 5/25.
The possible values are 15/25-10/25, 10/25-5/25 etc. Thus INSUFFICIENT.
Together, same thing. Possibility of white is 15/25, 10/25 etc and possibility of even is 10/25, 0/25 etc.
Thus, INSUFFICIENT.
So answer is E.
With no information, the number of white balls can be anything from 1-23 (assuming one is red and one blue).
The number of even can be anything from 0-25.
To calculate number of total balls which are either white or even, you should know, number of balls only white, number of balls only even and number of balls which are both even and white. The sum of these divided by 25 will give you the probability.
i) None of the white balls has an even number.
It is possible that there are 15 white balls and the rest 10 have even number, or it is possible that none of the 25 balls has even number. The answer can be 25/25=1, or 15/25, etc. INSUFFICIENT.
ii) Difference of probability of a white ball and an even numbered ball is 0.2 or 5/25.
The possible values are 15/25-10/25, 10/25-5/25 etc. Thus INSUFFICIENT.
Together, same thing. Possibility of white is 15/25, 10/25 etc and possibility of even is 10/25, 0/25 etc.
Thus, INSUFFICIENT.
So answer is E.
consider probability of getting a white ball as p(a)
consider probility of getting an even numbered ball as p(b)
then
probability of getting a white ball with an even number on it will be p(a^b)
probability of getting a white ball or an even numbered ball will be p(aUb)
by law
p(aUB) =p(a) +p(b) - p(a^b)
p(a^b) is given as 0
but
p(a)+p(b) is not given in any of the statements
instead p(a)-p(b) is given ,with which we can not get to a conclusion
therefore E should be the answer
please correct me if i am wrong
thanks
consider probility of getting an even numbered ball as p(b)
then
probability of getting a white ball with an even number on it will be p(a^b)
probability of getting a white ball or an even numbered ball will be p(aUb)
by law
p(aUB) =p(a) +p(b) - p(a^b)
p(a^b) is given as 0
but
p(a)+p(b) is not given in any of the statements
instead p(a)-p(b) is given ,with which we can not get to a conclusion
therefore E should be the answer
please correct me if i am wrong
thanks
