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# If x is to be chosen at random from the set {1, 2, 3, 4} and

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Intern
Joined: 02 Sep 2015
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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06 Oct 2015, 09:37
Gotcha! Thanks! I guess in this type of questions it's easiest to write out the possibilities?
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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26 Aug 2016, 09:04
I am poor at probability. Kindly explain where I went wrong:
If multiplication of numbers is even means at least one of the two numbers is even.

let a*b = even.
a is from first set. so probability of this being even: 2/4
b is from second set. so probability of this being even: 1/3
Accompanied by (or) so I added both.

I know this is wrong but am unable to figure out where I went wrong.
Kindly suggest a method in these lines.
Thanks
Intern
Joined: 10 Mar 2015
Posts: 41
Location: India
GMAT 1: 700 Q49 V37
GPA: 3.9
WE: Web Development (Computer Software)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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26 Aug 2016, 09:04
I am poor at probability. Kindly explain where I went wrong:
If multiplication of numbers is even means at least one of the two numbers is even.

let a*b = even.
a is from first set. so probability of this being even: 2/4
b is from second set. so probability of this being even: 1/3
Accompanied by (or) so I added both.

I know this is wrong but am unable to figure out where I went wrong.
Kindly suggest a method in these lines.
Thanks
Manager
Joined: 09 Aug 2016
Posts: 68
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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22 Oct 2016, 11:25
The best way to solve this is to indetify that the product of being even can be obtained by the following cases:

x=E * y=O = E
x=O * y=E = E
x=E * y=E = E

The equation above might look quite silly but effectivly when I am saying x=E * y=O I mean that x is even and y is odd. As soon as you find this relationship you did the hard part of the question.

Then is all about replacing the prob/ties:
x=E * y=O in GMAT terms can be written as x=E AND y=0 means P(x=E) AND P(y=O) = 2/4 * 2/3 = 4/12 = 1/3 etc...

Intern
Joined: 16 Oct 2017
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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21 Nov 2017, 12:11
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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02 Dec 2017, 18:14
Same question as above! A little confused why that doesn't work.

chetan2u can you please shed some light on this?

Thanks!

pablogutierrez24 wrote:
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?

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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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02 Dec 2017, 23:13
Same question as above! A little confused why that doesn't work.

chetan2u can you please shed some light on this?

Thanks!

pablogutierrez24 wrote:
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?

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
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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03 Dec 2017, 10:35
chetan2u wrote:
Same question as above! A little confused why that doesn't work.

chetan2u can you please shed some light on this?

Thanks!

pablogutierrez24 wrote:
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?

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

I guess I'm still a bit confused. For example, if a question was asking:

"A fair six-sided dice is rolled twice. What is the probability of getting at least once?"

Then I understand I could calculate the probability of not getting 3 twice and subtract: $$1 - (\frac{5}{6}*\frac{5}{6}) = \frac{11}{36}$$

Nevertheless, I could still say:
1. If I get 3 on my first roll I don't care what happen on the 2nd roll = $$\frac{1}{6}*\frac{6}{6}$$
2. If I don't get 3 on the 1st roll, I need 3 on the 2nd = $$\frac{5}{6}*\frac{1}{6}$$
3. Add the two results and get $$\frac{11}{36}$$

(Similarly, if the question was asking what is the probability that the sum of 2 rolls is 7, I would do $$1*\frac{1}{6}$$)

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'm thinking there is something about dependent / independent events I'm not completely getting...
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If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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11 Dec 2017, 16:50
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

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

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 =$$\frac{2}{4}*\frac{1}{3}$$
$$= \frac{2}{12} = \frac{1}{6}$$

$$(\frac{1}{2} + \frac{1}{6}) = (\frac{3}{6} + \frac{1}{6})= (\frac{4}{6}) = \frac{2}{3}$$

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?
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If x is to be chosen at random from the set {1, 2, 3, 4} and   [#permalink] 11 Dec 2017, 16:50

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