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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

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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

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.

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

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07 Feb 2014, 11:28

AKG1593 wrote:

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

I agree with you. But today, when I took the GMAT I encountered a similar question and my answer (according to that logic) wasn't there in the answer choices... So I assume it's not right.. I had x chosen at random from the numbers from 0 to 2 inclusive and y chosen at random from the numbers from 0 to 6 inclusive, what's the probability that x>y?

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

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07 Feb 2014, 11:42

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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.

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

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07 Feb 2014, 11:56

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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.

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

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18 Feb 2014, 20:19

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

The set {1, 2, 3, 4} contains 2 odd and 2 even numbers The set {5, 6, 7} contains 2 odd and 1 even numbers

Possible even xy = 2*2 + 2*2 = 8 and possible odd xy = 2*2 = 4

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

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07 Apr 2014, 03:46

arunspanda 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

I got it wrong the firs time, but 2nd time I used following method.

# of ways to select x = 4 # of ways to select y = 3 Total # of ways of getting xy = 4 * 3 = 12

If value of x is 1 then # of ways to get even value for xy = 1 (only 6 is possible value from set y) If value of x is 2 then # of ways to get even value for xy = 3 If value of x is 3 then # of ways to get even value for xy = 1 If value of x is 4 then # of ways to get even value for xy = 3

Total # of ways to get desired outcome = 1+3+1+3 = 8

Probability = desired outcome/ all possible outcomes = 8/12 = 2/3

Re: Probability- May i request you to help me in this one ? [#permalink]

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27 Sep 2014, 06:46

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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.

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

Sol: Atleast one = total - none Now for xy to be even, Atleast one even should be there between x and y Atleast one even = total - no even In terms of probability, prob(atleast one even)= probab(total) - probab(no even I.e. Odd) = 1 - 2/4 * 2/3 = 2/3 As we can choose 2 odds out of 4 from set 1 and 2 odds out of 3 from set 2
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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 May 2015, 00:04

shreyas wrote:

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.

In the case that both are even, shouldn't it matter which set we pick the even number from first?

Thus shouldn't the probability of picking 2 even numbers be: 1/2 * 1/3 * 2?

Are you considering only one case because the question stem asks about x*y only (and not y*x)?

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

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21 May 2015, 00:30

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avgroh wrote:

shreyas wrote:

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.

In the case that both are even, shouldn't it matter which set we pick the even number from first?

Thus shouldn't the probability of picking 2 even numbers be: 1/2 * 1/3 * 2?

Are you considering only one case because the question stem asks about x*y only (and not y*x)?

---short explanation-- question says probability that xy is even. Now xy = xy. So basically we need to find the probability of the product being even.

-- detailed explanation --for the case of both the numbers even the mathematical statement is simple-- one even number from set 1 AND one even number from set 2 (because we have to pic one number from each set). Order of picking is not relevant here. The question specifically says x is from first set and y is from the second set. The product has to be even and for a given value of (x,y), the product will be same whether you take product as xy or xy. The question is asking the probability of the "PRODUCT" being even.

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

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30 Jun 2015, 09:21

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.
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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 [#permalink]

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27 Sep 2015, 12:07

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?