Each of the 25 balls in a certain box is either red, blue or white and

Yes, we can conclude that white balls are only marked with odd numbers. But we have numbers on the balls from 1 to 10 only and we have total 25 balls. SO, it is obvious that numbers are repeated. Now, We don't have any information whether we do have only 5 ODD numbered white balls or more than 5 ODD numbered white balls. So, we cannot conclude the number of white balls. Hence, Statement 1 is also insufficient.

I hope it is clear now. Let me know if you have more doubts.

How do you get to this formula.:

\(P(WorE)=P(W)+P(E)-P(W&E)\)

Any underlying concept ?

Yes, it is a general formula to determine the probability of Events A or B when both A and B are NOT mutually exclusive.

For example, say I have the numbers from 1 to 10.

Event A : All even numbers
Event B : All multiples of 3.

Probability of A or B will be All even + All multiple of 3 - Those that have been repeated(only 6 in this case).

I hope it is clear now.

Exactly same as that of two overlapping sets.

n(A or B) = n(A) + n(B) - n(A and B)

Why do you subtract n(A and B) out of it? Because it is double counted - once in n(A) and another time in n(B).

Similarly, P(A and B) is double counted - once in probability of A and another time in probability of B.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

...........| White | Not White
------------------------------
.....Even.| a | b
---------------------------------
Not Even.| c | d

=> Q: ( a + b + c ) / 25 ?

a + b + c + d = 25

Since we have 4 variables and 1 equation, E could be the answer most likely.

1) & 2)
From the condition 1), we have a = 0.
From the condition 2), ( a + c ) / 25 - ( a + b ) / 25 = ( c - b ) / 25 = 0.2
c - b = 5

Since we have only 2 equations a = 0 and c - b = 5, we can't figure out b + c and solve this question.

Therefore, the answer is E as expected.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Hi Bunuel..

How did u get Probability ball picked being white or even: P(W or E)=P(W)+P(E)-P(W&E)?

OR probability:
If Events A and B are independent, the probability that Event A OR Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur: \(P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)\).

This is basically the same as 2 overlapping sets formula:
{total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B}.

Note that if event are mutually exclusive then \(P(A \ and \ B)=0\) and the formula simplifies to: \(P(A \ or \ B) = P(A) + P(B)\).

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: \(P(A \ and \ B) = P(A)*P(B)\).

This is basically the same as Principle of Multiplication: if one event can occur in \(m\) ways and a second can occur independently of the first in \(n\) ways, then the two events can occur in \(mn\) ways.

22. Probability

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  GMAT Tip of the Week: 99 Problems But Probability Ain't One
  
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    Advanced Topic: Probability and Combinations Questions With Solutions
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Hope it helps.

Can someone tell me where my logic went wrong? I selected A, with the thought that since P(W&E)=0, therefore there must be 5 white balls, all of which have an odd number on them (so 5 white balls total). Therefore, P(W)=5/25, P(E)=10/25, and P(W&E)=0. Is it wrong to assume that the numbers don't repeat on each ball color (i.e. the blue balls with all of 1 number cannot be true?).

Hi

P(W&E) = 0, means that there is NO ball which is both white and has an even number on it. But there could be various white balls with odd numbers on them, and there could be various red/blue balls with even numbers on them. We need to take both these kinds of balls into account.

I think we cannot assume that there must be 5 white balls, its possible that there is only one white ball in the entire box. And yes, I think we also cannot assume that the numbers cannot repeat on same coloured balls (because its nowhere given). Its possible that all blue coloured balls have only number 1 painted on them.

Hi Bunuel,

This is an interesting approach to this problem.

Please let me know when can we use this Transformation of probability into actual numbers.

And also if you have any similar questions in which I can apply this technique for practice.

Thanks in advance!!

PS Overlapping + Probability questions

DS Overlapping + Probability questions

Hope it helps.

Correct Option : E

the best part of DS is, one dont have to dig down the answer and logically to answer in few allocated mins.
either it will be YES or NO - simple, that's what GMAT test in such Question

to solve, we need 4 parameter
1. P(white or even)
2. P(white)
3. P(even)
4. P(white & even)

is it availabe in
A - YES or NO, No then Eliminate
B - YES or NO, No then Eliminate
C - by both (A & B)YES or NO, No then Eliminate
D - Is It not either A or B - YES or NO, No then Eliminate
E - No solution - WINNER

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