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

[RonPurewal]

I just solved this problem using properties of sets:
P(A) or P(B) = P(A) + P(B) - P(both A&B)

We are told in S1: P(A&B) = 0, we still need P(A) and p(B), so S1 is insufficient

S2: gives us: P(A) - P(B), but we need P(A)+(B), so S2 is also insufficient. from s2, A and B can take multiple values.

Combining both S1 and S2 does not still provide p(A)+p(B), hence
answer is E.

Note: A stands for white, B is Even number

this looks legitimate. however, note that it's not as generalizable as the overlapping-sets methods.

still, more methods = better, so it's good to have this approach as well.

[winstondkim]

let,
p(W) = probability the ball is White
p(E) = probability that the ball is painted with even number on it.

we can manipulate both of given conditions algebraically to produce p(W)+p(E), can't we?

1) P(W) x P(E) = 0
2) P(W) - P(E) = 0.2

if we square 2),
P(W)^2 + P(E)^2 - 2p(W)p(E) = 0.04

since p(W)p(E) = 0,
P(W)^2 + P(E)^2 - 2P(W)P(E) + 4p(W)p(E) = 0.04
{P(W)+P(E)}^2 = 0.04
P(W)+P(E) = 0.2

please comment on this. thanks.

[RonPurewal]

1) P(W) x P(E) = 0

this is incorrect. the statement
prob(w AND e) = 0
is definitely not the same as
prob(w) x prob(e) = 0.

there are certain conditions under which this statement is true, but those conditions are not tested on the gmat exam. therefore, the only thing you have to know about this idea is that it's false.

it's easy to demonstrate with examples. let's say i have a room of people. there are men and women. of the men, some have red hair and some have dark hair. however, all of the women have dark hair.
in this situation:
prob(female AND red hair) = 0, since there are no red-headed women in the room.
but...
prob(female) x prob(red hair) is not 0, since there are both women and redheads (= red-headed males, in this case) in the room.

always check your principles with easy examples before you try to use them!

[griffin.811]

Is the "property of sets" discussed in the manhattan guides? I just took a look at the number properties guide and didn't see anything.

Possibly called something else?

Thanks

[RonPurewal]

Is the "property of sets" discussed in the manhattan guides? I just took a look at the number properties guide and didn't see anything.

Possibly called something else?

Thanks

I have no idea what this is. And it doesn't seem to have been mentioned anywhere in this thread, either.

I'm confused.

Context?

Source?

[Ace1]

Hi, I tried to solve st 2 as shown below:
P(white) - P(even) = 0.2
P(even)=12/25 ('coz there are 12 even nos in the box)
so, P(white)= 0.2 + 12/25 = 17/25
P(white) + P (even) = 12/25 + 17/25 = 1.16 which is incorrect and hence B is out.

Is the above understanding correct?

[RonPurewal]

Hi, I tried to solve st 2 as shown below:
P(white) - P(even) = 0.2
P(even)=12/25 ('coz there are 12 even nos in the box)
so, P(white)= 0.2 + 12/25 = 17/25
P(white) + P (even) = 12/25 + 17/25 = 1.16 which is incorrect and hence B is out.

Is the above understanding correct?

you can't just add the probabilities together. that only works if there is no overlap.
if there are any balls that are both white and even, then you're double-counting those.

[RonPurewal]

also, more fundamentally--
it seems you don't have a solid understanding of how data sufficiency works.

there's no such thing as an "incorrect" statement in DS.

every statement will have at least one solution.
if there's only one answer, then the statement is "sufficient"; if there are two or more, then "not sufficient".

so, if you think that a statement is impossible / is "incorrect" / has no solution ... you're doing something wrong.

[sahilk47]

Hi, I tried to solve st 2 as shown below:
P(white) - P(even) = 0.2
P(even)=12/25 ('coz there are 12 even nos in the box)
so, P(white)= 0.2 + 12/25 = 17/25
P(white) + P (even) = 12/25 + 17/25 = 1.16 which is incorrect and hence B is out.

Is the above understanding correct?

you can't just add the probabilities together. that only works if there is no overlap.
if there are any balls that are both white and even, then you're double-counting those.

Hi Ron

I was going through the entire thread to understand better the question and your response to different posts. I have one doubt. In the post above your's, Ace1 has written P(even) = 12/ 25 because there are 12 even nos. How do we know that there are 12 balls ( and not any other number of balls) with even number on them ?

Thank you!

[RonPurewal]

hm. no, that isn't something we know.

perhaps that poster (mistakenly) thought that the balls were painted with the integers 1, 2, 3, ..., 25. in that case there would be twelve evens and thirteen odds.
...but in fact we know that's not the case, because no ball is marked with any integer greater than 10.

(that poster's method was already wrong, so i don't even think i bothered to look at the actual numbers that (s)he used.)

[sahilk47]

hm. no, that isn't something we know.

perhaps that poster (mistakenly) thought that the balls were painted with the integers 1, 2, 3, ..., 25. in that case there would be twelve evens and thirteen odds.
...but in fact we know that's not the case, because no ball is marked with any integer greater than 10.

(that poster's method was already wrong, so i don't even think i bothered to look at the actual numbers that (s)he used.)

Thank you for the clarification!

[RonPurewal]

no problem.

thanks to you, too, for raising the issue (since i didn't catch it earlier).