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

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

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.

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

Add 1+2+3 ==> 2/3 ----------D

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!

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

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.

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

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/

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

Answer: D

Cheers,
Brent

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)

Devmitra Sen
GMAT SME

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 - \(\frac{1}{3}\) = \(\frac{3 - 1}{3}\) = \(\frac{2}{3}\)

So, Answer will be D.
Hope it helps!

@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.

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

please clarify ?

thanks in advance.

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,

genuinely need your further clarify.
thanks a lot