probability that xy will be even

This topic has expert replies
User avatar
Legendary Member
Posts: 1325
Joined: Sun Nov 01, 2009 6:24 am
Thanked: 105 times
Followed by:14 members

probability that xy will be even

by vikram4689 » Sun Mar 11, 2012 2:02 am
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


XY will be even if one of the 2 no.'s is even.
P(even from set 1)=2/4
P(even from set 2)=1/3
P(XY is even) = 2/4 + 1/3 =5/6... What is WRONG here
Premise: If you like my post
Conclusion : Press the Thanks Button ;)

Legendary Member
Posts: 581
Joined: Sun Apr 03, 2011 7:53 am
Thanked: 52 times
Followed by:5 members

by killer1387 » Sun Mar 11, 2012 2:15 am
vikram4689 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


XY will be even if one of the 2 no.'s is even.
P(even from set 1)=2/4
P(even from set 2)=1/3
P(XY is even) = 2/4 + 1/3 =5/6... What is WRONG here
xy will be even if
1. both x &y even
2. one of them is even and other is odd.
i.e.(x,y)= (e,o)+(o,e)+(e,e)

REQUIRED PROBABILITY= 1- (BOTH X &Y ODD)= 1-(2/4*2/3)= 8/12=2/3
hence D.

HTH.

Senior | Next Rank: 100 Posts
Posts: 99
Joined: Sun Mar 13, 2011 12:42 pm

by factor26 » Sun Mar 11, 2012 7:27 am
Since the sampling area ( x and y ) is rather small we can map out all the instances when X*Y will yield a even integer.

X {1,2,3,4}
Y {5,6,7}

X=1 Y=6
X=2 Y=5 Y=6 OR Y=7
X=3 Y=6
X=4 Y=5 Y=6 Y=7

TOTAL # OF MAKING EVEN #'S = 8

TOTAL # OF CHOICES OR OUTCOMES 4 * 3 = 12

= 8/12
=4/6
=2/3

THUS THE ANSWER IS D

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Mar 11, 2012 8:04 am
vikram4689 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
Another 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) times P(y is odd)
= (2/4) (2/3)
= 1/3


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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

User avatar
Legendary Member
Posts: 1325
Joined: Sun Nov 01, 2009 6:24 am
Thanked: 105 times
Followed by:14 members

by vikram4689 » Sun Mar 11, 2012 9:58 pm
Brent what is wrong in method that i used
Premise: If you like my post
Conclusion : Press the Thanks Button ;)

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Mon Mar 12, 2012 2:58 am
vikram4689 wrote:Brent what is wrong in method that i used
Here's your solution:

XY will be even if one of the 2 no.'s is even.
P(even from set 1)=2/4
P(even from set 2)=1/3
P(XY is even) = 2/4 + 1/3 =5/6


Here, you have said that xy will be even if one of the 2 no.'s is even. Great!
In other words, P(xy is even) = P(x is even or y is even)

For "or" probabilities, the formula is: P(A or B) = P(A) + P(B) - P(A and B) [you missed the P(A and B) part]

So, P(x is even or y is even) = P(x is even) + P(y is even) - P(x is even and y is even)
= (2/4) + (1/3) - [(2/4)(1/3)]
= 2/3
= D


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

User avatar
Legendary Member
Posts: 1325
Joined: Sun Nov 01, 2009 6:24 am
Thanked: 105 times
Followed by:14 members

by vikram4689 » Mon Mar 12, 2012 9:08 am
Thanks Brent, arrrrghh me for such a silly mistake
Premise: If you like my post
Conclusion : Press the Thanks Button ;)

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1462
Joined: Thu Apr 09, 2015 9:34 am
Location: New York, NY
Thanked: 39 times
Followed by:22 members

hi

by Jeff@TargetTestPrep » Wed Dec 13, 2017 10:12 am
vikram4689 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 1/3; 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 2/3; 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 1/2, and the probability that y is even is 1/3; 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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews