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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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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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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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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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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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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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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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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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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]
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
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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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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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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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Bunuel 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

One approach is to recognize that P(xy is even) = 1 - P(xy is not even)
In other words, P(xy is even) = 1 - P(xy is odd)

Aside: This is a useful approach since there's only one way that xy can be odd. Both x and y must be odd for their product to be odd.
Conversely, there are 3 different cases to consider for xy to be even: 1) x and y are both even. 2) x is odd and y is even. 3) x is even and y is odd.

P(xy is odd) = P(x is odd AND y is odd)
= P(x is odd) x P(y is odd)
= 2/4 x 2/3
= 1/3

So, P(xy is even) = 1 - 1/3
= 2/3

Cheers,
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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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Solution:

Probability of choosing an odd number from set X = 2/4

Probability of choosing an odd number from set Y = 2/3

The possibilities for choosing a number from set X that is Odd and another from set Y that is also odd

= 2/4 * 2/3

= 1/3

Now, 1- 1/3 = 2/3 is the probability of choosing two numbers which can be even, odd or even, even or odd, even implying the product xy will always be even.

(option 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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Given that 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} and we need to find what is the probability that xy will be even

We know that Total Probability is always equal to 1

=> P(Odd) + P(Even) = 1
=> P(Even) = 1 - P(Odd)

To get xy as odd we need to have all of them as odd.
=> P(xy = Odd) = P(x = odd) * P(y = odd) = $$\frac{2}{4}$$ * $$\frac{2}{3}$$

(Because for x we have two choices (1,3) to get odd out of 4, for y we have two choices (5,7) to get odd out of 3)

= $$\frac{1}{3}$$

=> P(Even) = 1 - P(Odd) = 1 - $$\frac{1}{3}$$ = $$\frac{3 - 1}{3}$$ = $$\frac{2}{3}$$

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]
@bunuel/@chetan2u/@KarishmaB

I have understood the logic of the question, But how you have arrived at the conclusion that 'xy' is a product and not a random number from Set A and Set B. (for eg. 15, 16, 17) because in that case, probabality would have been 1/3.

Kindly can you help ?
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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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SnorLax_7 wrote:
@bunuel/@chetan2u/@KarishmaB

I have understood the logic of the question, But how you have arrived at the conclusion that 'xy' is a product and not a random number from Set A and Set B. (for eg. 15, 16, 17) because in that case, probabality would have been 1/3.

Kindly can you help ?

xy there means x*y, x multiplied by y not a two-digit integer xy, recall that multiplication sign is often omitted. If xy were a two-digit number, it would have been mentioned explicitly.
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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]
JeffTargetTestPrep wrote:
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.

hi JeffTargetTestPrep
I did not consider case 1, because I thought case 2 &3 include case 1, leading me to pick up E

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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]
zoezhuyan wrote:
JeffTargetTestPrep wrote:
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.

hi JeffTargetTestPrep
I did not consider case 1, because I thought case 2 &3 include case 1, leading me to pick up E

There are four complementary, and thus mutually exclusive, scenarios when choosing two numbers: EE, EO, OE, and OO.

• EE: Probability is 1/6
• EO: Probability is 1/3
• OE: Probability is 1/6
• OO: Probability is 1/3

The total probability adds up to 1 (1/6 + 1/3 + 1/6 + 1/3 = 1). The first three scenarios (EE, EO, OE) result in an even product xy. Therefore, scenarios 2 and 3 do not include scenario 1.
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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]
Bunuel wrote:
There are four complementary, and thus mutually exclusive, scenarios when choosing two numbers: EE, EO, OE, and OO.

• EE: Probability is 1/6
• EO: Probability is 1/3
• OE: Probability is 1/6
• OO: Probability is 1/3

The total probability adds up to 1 (1/6 + 1/3 + 1/6 + 1/3 = 1). The first three scenarios (EE, EO, OE) result in an even product xy. Therefore, scenarios 2 and 3 do not include scenario 1.

hi Bunuel
say case 2, x=even, y = whatever
then x is selected from 2 or 4,
y selected from 5,6,7 -- so it is highly possible to picked up x=2, y=6

say case 1, both x and y are even
so it is possible x=2,y=6,

you seem case 1 is already included in case 2

that's why I didn't consider case1,