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

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

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05 Feb 2014, 01:29
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35% (medium)

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67% (01:06) correct 33% (00:59) wrong based on 1452 sessions

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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

Problem Solving
Question: 79
Category: Arithmetic; Algebra Probability; Concepts of sets
Page: 72
Difficulty: 600

The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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05 Feb 2014, 01:29
8
6
SOLUTION

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

Probably the best way to solve would be to use 1 - P(opposite event) = 1 - P(odd) = 1 - P(odd)*P(odd) = 1 - 2/4*2/3 = 8/12 = 2/3.

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

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07 Feb 2014, 11:56
7
2
At least one of the numbers must be even.
Both even: 1/2 * 1/3 = 1/6.
At least one even: even from first set and odd from second set: 1/2 * 2/3 = 1/3
At least one even: odd from first set and even from second set: 1/2 * 1/3 = 1/6
Reqd prob = 1/6 + 1/3 + 1/6 = 2/3.
##### General Discussion
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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05 Feb 2014, 07:36
2
Ans D

In order to make even nos. by multiplication,
we should have even * odd, odd * even or even * even

Total even nos. possible by multiplying nos. from the 2 sets:
(1*6);
2* any of the three from Set B;
3*6;&
4* any of the three from set B
1+3+1+3=8

Total possibilities=4 * 3=12

P(event)=8/12 or 2/3
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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07 Feb 2014, 11:42
1
x*y is even only when they're both even or one of them is odd. The probability of any single outcome is given by 1/3*1/4 = 1/12.
Working through possible outcomes, we arrive at 8/12 -> 2/3, hence (D).

It took me around a minute and 40 seconds to solve this, I'm sure that a faster approach exists, waiting for others.
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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07 Feb 2014, 20:58
5
3 Scenario's

E * E = 2/4 * 1/3 = 1/6 ----- 1

E * O= 2/4 * 2/3 = 4/12-----2

O* E = 2/4 *1/3 = 2/12------3

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

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27 Sep 2014, 06:46
1
xy will be even when either x or y or both are even.
Let's consider both x and y are old, so the product will be odd, not even. How many values can x take? There are only two odd values in the given set. Likewise, there are only two odd values possible for y. Since we want x and y to be odd, the total possible odd values for xy is 2*2=4.
Total number of possible xy values is 4*3=12 (x can take any 4 values from the set and y can take any 3 values from the set). So probability of xy to be odd is 4/12=1/3. So probability of xy to be even is 1-(probability of being odd) = 1-1/3 = 2/3.
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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30 Jun 2015, 09:34
BrainLab wrote:
Why do we count here only possibilities from Set 1 to Set 2 and not also vice versa as with this example

Two number cubes with faces numbered 1 to 6 are rolled. What is the probability
that the sum of the rolls is 8?

Here are the rolls that work:
2 - 6 3 - 5 4 - 4 5 - 3 6 - 2
That’s it; there are 5 combinations that work. Therefore the probability of a sum of 8 is 5/36.

The example that you have quoted here is the example of arrangements

When we say total Outcomes of two rolls = 6 x 6 = 36 [It includes all arrangements like 1-6 and 6-1 separately]

However, when we calculate the Products the the arrangement doesn't matter 2*3 is same as 3*2 because the result is 6 which is identical

When we say total Outcomes in this case = 4 * 3 = 12 [It considers all possible multiplication just once]

1x5
1x6
1x7
2x5
2x6
2x7

3x5
3x6
3x7
4x5
4x6
4x7

So probability = 8/12 = 2/3

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

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23 Jul 2015, 05:47
2
2
jimmyjamesdonkey wrote:
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

Method-1
xy will be even when
1) x is even and y is odd, Probability of x even is (2/4) and Probability of y odd is (2/3), so Probability of Case(1) = (2/4)*(2/3) = (4/12)
2) x is Odd and y is Even, Probability of x Odd is (2/4) and Probability of y Even is (1/3), so Probability of Case(2) = (2/4)*(1/3) = (2/12)
3) x is even and y is Even, Probability of x even is (2/4) and Probability of y Even is (1/3), so Probability of Case(1) = (2/4)*(1/3) = (2/12)

Total Favorable Probability = (4/12)+(2/12)+(2/12) = (8/12) = 2/3

Method-2
Unfavorable case is when x and y are both odd,

Unfavorable Probability = (2/4)*(2/3) = 1/3

i.e. favorable probability = 1-(1/3) = 2/3

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

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23 Jul 2015, 06:30
1
Let A = {1,2,3,4}, B = {5,6,7}.

Total mappings from A->B = 4 * 3 = 12.

Number of mappings from A->B where A = {2,4} and B = {6}: 2 * 1 = 2 [even -> even]
Number of mappings from A->B where A = {1,3} and B = {6}: 2 * 1 = 2 [odd -> even]
Number of mappings from A->B where A = {2,4} and B = {5,7}: 2 * 2 = 4 [even -> odd]

Total favourable events = 2+2+4 = 8.

Probability = 8/12 = 2/3. Ans (D).
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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29 Sep 2015, 22:15
If you write the cases, they are {(2,5) (2,6) (2,7)}, {(4,5),(4,6),(4,7)} & {(6,1),(6,2),(6,3),(6,4)}

As you can observe, there are two common cases out of total options of 12. So, probability = 2/12=1/6 needs to subtracted from the addition .

So, 1/2+1/3-1/6 = 4/6 =2/3

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

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08 Dec 2015, 21:14
To avoid double counting it is better to just count one way. For instance,
1) You choose 2 0r 4 from set1 and any number from set2: probability = 1/2*1
2) You choose 1 or 3 from set1 and 6 from set2: Probability = 1/2*1/3= 1/6
Total Probability= 1/2+1/6= 8/12 = 2/3

Hey all, I'm having trouble understanding why my approach doesn't work and I'd really appreciate your help! I picked E because for xy to be even, you need to pick an even number in first set (and you don't care about what you pick in second set), or an even number in second set (and you don't care what you pick in first set). So then the probability of even in first set is 1/2 and second set is 1/3. Why can't I just add the probabilities together to have 5/6? What have I included in the probability to make it 1/6 bigger than the right answer?

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

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27 Mar 2017, 10:49
jimmyjamesdonkey wrote:
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

In order for xy to be even, at least one of the values of x and y needs to be even. We know that (1) even x even = even and (2) even x odd = even, and (3) odd x even = even.

Case 1. Both x and y are even. The probability that x is even is 2/4 = 1/2 , and the probability that y is even is ⅓; thus, the probability that x and y will both be even is 1/2 x 1/3 = 1/6

Case 2. x is even and y is odd. The probability that x is even is 1/2, and the probability that y is odd is ⅔; thus, the probability that x is even and y is odd is 1/2 x 2/3 = 2/6 = 1/3.

Case 3. x is odd and y is even. The probability that x is odd is ½, and the probability that y is even is ⅓; thus, the probability that x is odd and y is even is 1/2 x 1/3 = 1/6.

Thus, the total probability that the product xy will be even is 1/6 + 1/3 + 1/6 = 1/6 + 2/6 + 1/6 = 4/6 = 2/3.

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

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18 Sep 2017, 05:48
A noob question, if one set gives even does other set even matter as product would be even.

Probability to get even out of 1,2,3,4 - Is 2/4 or 1/2

Probability to get even out of 5,6,7 - Is 1/3

Shouldnt it be then

Pset1(even)*1(as other set doesn't matter-answer would be even) + Pset2(even)*1 =1/2+1/3.

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

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18 Sep 2017, 21:10
1
Anantz wrote:
A noob question, if one set gives even does other set even matter as product would be even.

Probability to get even out of 1,2,3,4 - Is 2/4 or 1/2

Probability to get even out of 5,6,7 - Is 1/3

Shouldnt it be then

Pset1(even)*1(as other set doesn't matter-answer would be even) + Pset2(even)*1 =1/2+1/3.

Posted from my mobile device

Here is the problem with this solution:

Pset1(even)*1 includes the probability that both elements are even.
Pset2(even)*1 also includes the probability that both elements are even.

So a case such as 2, 6 is counted twice. We need to remove once the probability that both are even.
Pset1(even)*Pset2(even) = (1/2)*(1/3) = 1/6

So to get the answer, 1/2 + 1/3 - 1/6 = 2/3

I have discussed the same thing here:
https://gmatclub.com/forum/if-x-is-to-b ... l#p1321459

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

You can also use the sets formula:

P(xy is even) = P(x is even) + P(y is even) - P(Both x and y are even) (We subtract this to get rid of the double counting)
P(xy is even) = 2/4 + 1/3 - (2/4)(1/3) = 2/3

You can do many questions using the same concept. Check out this post for more on this: http://www.veritasprep.com/blog/2012/01 ... e-couples/
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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21 Nov 2017, 11:11
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 y is to  [#permalink]

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02 Dec 2017, 17:14
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 y is to  [#permalink]

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02 Dec 2017, 22:13
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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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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03 Dec 2017, 09:35
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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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to  [#permalink]

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11 Dec 2017, 15:50
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?
_________________

Never look down on anybody unless you're helping them up.
--Jesse Jackson

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to &nbs [#permalink] 11 Dec 2017, 15:50

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