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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
Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.
We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
05 Feb 2014, 01:29
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Expert's post
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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.
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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]
07 Feb 2014, 11:42
1
This post received KUDOS
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]
07 Feb 2014, 11:56
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This post received KUDOS
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]
08 Feb 2014, 03:08
Expert's post
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.
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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]
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]
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 _________________
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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]
21 May 2015, 00:30
1
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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]
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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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
30 Jun 2015, 09:34
Expert's post
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]
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