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
probability that xy will be even
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xy will be even ifvikram4689 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
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
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
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Another approach is to recognize that P(xy is even) = 1 - P(xy is not even)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 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
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Brent what is wrong in method that i used
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Here's your solution:vikram4689 wrote:Brent what is wrong in method that i used
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
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Thanks Brent, arrrrghh me for such a silly mistake
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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.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
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
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