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

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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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New post Updated on: 17 Sep 2013, 07:23
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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

Originally posted by jimmyjamesdonkey on 16 Jul 2007, 15:25.
Last edited by Bunuel on 17 Sep 2013, 07:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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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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New post 23 Jan 2014, 21:02
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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


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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New post 16 Jul 2007, 16:51
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OA is 2/3. Good job all. I actually got this wrong, but here is a great tip. I think everytime you see a probability problem, look at what it is asking, and then find the opposite of it.

For example, it is asking you to find the prob of xy being even...instead find the prob of it being ODD.

How is it odd? If x and y are both ODD.

Prob of x being odd: 1/2
prob of y being odd: 2/3
Prob of x and y being ODD: 1/2 * 2/3 = 1/3

Now 1 - 1/3 = 2/3..our answer.

I find that on all the OG problems dealing with prob, looking at it in reverse seems to bring you to a answer quicker. Anyone agree with that?
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New post 16 Jul 2007, 15:49
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it should be D
even x even =1/2*1/3
evenx odd= 1/2* 2/3
odd*even= 1/2*1/3

P= 1/6+ 2/6+ 1/6= 2/3
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Re: PS - Probability  [#permalink]

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New post 16 Jul 2007, 16:03
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jimmyjamesdonkey wrote:
Q: 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


there are 12 different possibilities:
4C1 = 4
3C1 = 3
4(3) = 12

there's a 2/3 chance of set {5,6,7} giving us an odd # and there is a 1/2 chance of set {1,2,3,4} giving us an odd#. if only odd * odd = odd, there is a 1/3 = (2/3)*(1/2) chance of getting an odd product. therefore, there's a 8/12 ((1/3)*12 = 4) chance of getting an even product. 8/12 = 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  [#permalink]

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New post 23 Jan 2014, 14:01
I think we can use combinations too, to arrive at the answer.

The total number of combinations is 12 (4C1 for X * 3C1 for Y). The possibilities for having XY as even are:

1) when X is 1, Y has to be 6. That makes it 1C1*1C1
2) when X is 2, Y can take 5,6,7. This would be 1C1*3C1
3) when X is 3, Y can take only 6. So, this would mean 1C1*1C1
4) when X is 4, Y can take 5,6,7. That would be 1C1*3C1

So, 1C1*1C1 + 1C1*3C1 + 1C1*1C1 + 1C1*3C1 is divided by 4C1*3C1, i.e when 1 + 3 +1 + 3 is divided by 4*3 we get 2/3.

Please let me know if I am wrong.
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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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New post 23 Jul 2015, 06:47
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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

Answer: option D

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

Answer: option D
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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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New post 23 Jul 2015, 07:30
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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  [#permalink]

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New post 05 Oct 2015, 01:33
Set A={1, 2, 3, 4}
Set B={5, 6, 7}
For the product xy to be even , atleast one of the selected numbers should be even .
Method 1 = Prob(x is even and y is even) + Prob(x is even and y is odd) + Prob(x is odd and y is even)

Method 2=Prob(Favourable outcomes)= 1 - Prob(Unfavourable outcomes) =1- Prob(x is Odd and y is Odd)

The method of finding the unfavourable outcomes and subtracting the result from 1 will be quicker .

:)
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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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New post 23 Mar 2017, 05:02
my approach was to start apply 1-xy (odd). As odd multiply can be only in case x and y are odd.

X set: 2/4
Y set: 2/3
2/4 and 2/3 = 4/12
1- 4/12=8/12=2/3
Answer is D

Another approach:
e*o=2/4*2/3
o*e=2/4*1/3
e*e=2/4*1/3
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  [#permalink]

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New post 27 Mar 2017, 11:49
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.

Answer: D
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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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New post 18 Sep 2017, 06:48
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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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

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New post 18 Sep 2017, 22:10
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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
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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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New post 18 Sep 2017, 22:26
2&6 and 6&2 getting in twice.


Got it thank you.

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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and &nbs [#permalink] 18 Sep 2017, 22:26
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