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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

Expert's post

2

This post was BOOKMARKED

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

2

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

Probability- May i request you to help me in this one ? [#permalink]
27 Sep 2014, 06:34

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

Re: Probability- May i request you to help me in this one ? [#permalink]
27 Sep 2014, 06:46

1

This post received KUDOS

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.

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
27 Sep 2014, 11:42

Expert's post

enggag 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

Merging similar topics. Please refer to the discussion above.

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