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.

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

Brent what is wrong in method that i used

Thanks Brent, arrrrghh me for such a silly mistake

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