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If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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16 Jul 2007, 15:25

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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?

Q: 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

there are 12 different possibilities:
4C1 = 4
3C1 = 3
4(3) = 12

there's a 2/3 chance of set {5,6,7} giving us an odd # and there is a 1/2 chance of set {1,2,3,4} giving us an odd#. if only odd * odd = odd, there is a 1/3 = (2/3)*(1/2) chance of getting an odd product. therefore, there's a 8/12 ((1/3)*12 = 4) chance of getting an even product. 8/12 = 4/6 = 2/3

OA is 2/3. Good job all. I actually got this wrong, but here is a great tip. I think everytime you see a probability problem, look at what it is asking, and then find the opposite of it.

For example, it is asking you to find the prob of xy being even...instead find the prob of it being ODD.

How is it odd? If x and y are both ODD.

Prob of x being odd: 1/2
prob of y being odd: 2/3
Prob of x and y being ODD: 1/2 * 2/3 = 1/3

Now 1 - 1/3 = 2/3..our answer.

I find that on all the OG problems dealing with prob, looking at it in reverse seems to bring you to a answer quicker. Anyone agree with that?

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

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23 Jan 2014, 14:01

I think we can use combinations too, to arrive at the answer.

The total number of combinations is 12 (4C1 for X * 3C1 for Y). The possibilities for having XY as even are:

1) when X is 1, Y has to be 6. That makes it 1C1*1C1 2) when X is 2, Y can take 5,6,7. This would be 1C1*3C1 3) when X is 3, Y can take only 6. So, this would mean 1C1*1C1 4) when X is 4, Y can take 5,6,7. That would be 1C1*3C1

So, 1C1*1C1 + 1C1*3C1 + 1C1*1C1 + 1C1*3C1 is divided by 4C1*3C1, i.e when 1 + 3 +1 + 3 is divided by 4*3 we get 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

You can also use the sets formula:

P(xy is even) = P(x is even) + P(y is even) - P(Both x and y are even) (We subtract this to get rid of the double counting) P(xy is even) = 2/4 + 1/3 - (2/4)(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

Method-1 xy will be even when 1) x is even and y is odd, Probability of x even is (2/4) and Probability of y odd is (2/3), so Probability of Case(1) = (2/4)*(2/3) = (4/12) 2) x is Odd and y is Even, Probability of x Odd is (2/4) and Probability of y Even is (1/3), so Probability of Case(2) = (2/4)*(1/3) = (2/12) 3) x is even and y is Even, Probability of x even is (2/4) and Probability of y Even is (1/3), so Probability of Case(1) = (2/4)*(1/3) = (2/12)

Total Favorable Probability = (4/12)+(2/12)+(2/12) = (8/12) = 2/3

Answer: option D

Method-2 Unfavorable case is when x and y are both odd,

Unfavorable Probability = (2/4)*(2/3) = 1/3

i.e. favorable probability = 1-(1/3) = 2/3

Answer: option D
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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23 Jul 2015, 07:30

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Let A = {1,2,3,4}, B = {5,6,7}.

Total mappings from A->B = 4 * 3 = 12.

Number of mappings from A->B where A = {2,4} and B = {6}: 2 * 1 = 2 [even -> even] Number of mappings from A->B where A = {1,3} and B = {6}: 2 * 1 = 2 [odd -> even] Number of mappings from A->B where A = {2,4} and B = {5,7}: 2 * 2 = 4 [even -> odd]

Total favourable events = 2+2+4 = 8.

Probability = 8/12 = 2/3. Ans (D).
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

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05 Oct 2015, 01:33

Set A={1, 2, 3, 4} Set B={5, 6, 7} For the product xy to be even , atleast one of the selected numbers should be even . Method 1 = Prob(x is even and y is even) + Prob(x is even and y is odd) + Prob(x is odd and y is even)

Method 2=Prob(Favourable outcomes)= 1 - Prob(Unfavourable outcomes) =1- Prob(x is Odd and y is Odd)

The method of finding the unfavourable outcomes and subtracting the result from 1 will be quicker .

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

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20 Oct 2016, 03:01

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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

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 ⅓; 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 ⅔; 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 ½, and the probability that y is even is ⅓; 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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