Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition
If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6
Hadrienlbb wrote:
chetan2u wrote:
Hi...
What you are doing here is " finding probability of picking an even number from atleast one of the set"
Say the set B were 5,6,8 so two even
Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1?
But we are looking for xy to be even..
This will include following cases..
1) even from A and odd number from B...
2/4 * 2/3=1/3
2) even from A and even from B
2/4 *1/3=1/6
3) odd from A and even from B
2/4*1/3=1/6
Total =1/3+1/6+1/6=2/3
Other easier way would be -
Subtract prob of odd from 1
Prob of odd .. odd from both = 2/4*2/3=1/3
So Ans=1 -1/3=2/3
Thanks for your response!
I guess I'm still a bit confused.
.
.
.
So here I went about calculating:
1. If I get even from the first set I don't care what the second set gets me =
\(\frac{2}{4}*1\)2. If I get an odd from the first set, I need an even from the second set =\(\frac{2}{4}*\frac{1}{3}\)
Now I understand this is wrong, and my first extinct would to subtract the probability of getting two odds.
I just don't really understand why this is wrong lol. [I don't think it IS wrong.] I'm thinking there is something about dependent / independent events I'm not completely getting...
Hadrienlbb , I think the part in red is not accurate.
I am not sure where you think your mistake is, because if I finish your math, the answer is correct.
Here is the math from your steps, finished:
Quote:
1. If I get even from the first set I don't care what the second set gets me = \(\frac{2}{4}*1\)
\(= \frac{1}{2}\)
2. If I get an odd from the first set, I need an even from the second set =\(\frac{2}{4}*\frac{1}{3}\)
\(= \frac{2}{12} = \frac{1}{6}\)
Now add the results of your numbers 1 and 2:
\((\frac{1}{2} + \frac{1}{6}) = (\frac{3}{6} + \frac{1}{6})= (\frac{4}{6}) = \frac{2}{3}\)
You just compressed
chetan2u 's first two steps, which, with the third (and your second), he adds.
Here are his first two of three steps:
1) even from A and odd number from B...
2/4 * 2/3=1/3
2) even from A and even from B
2/4 *1/3=1/6
REWRITE:
(2/4 *
2/3) + (2/4 *
1/3) =
2/4 *
(2/3 + 1/3) =
2/4 *
(1) -- which is exactly what you calculated.
That's the result for picking an even number from set A.
Then add the result for picking an odd number from set A (which both you and he computed identically)
\((2/4 * 1) + \frac{1}{6} =\)
\((\frac{1}{2} + \frac{1}{6}) = \frac{6 + 2}{12}= \frac{8}{12} =\frac{2}{3}\)
What you did here is not the same as the person you were quoting . . . (and I had to delete all that because the machine hollers when there are too many quotes within quotes).
Maybe I am missing something. Now I am the confused one, I think.
It seems to me you got it right.
Why do you think there is a mistake?
_________________
The humblest praise most, while cranks & malcontents praise least. Praise almost seems to be inner health made audible. ~ C. S. Lewis