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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to

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Bunuel
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink] New post   30 Nov 2023, 05:38
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zoezhuyan wrote:
Bunuel wrote:
There are four complementary, and thus mutually exclusive, scenarios when choosing two numbers: EE, EO, OE, and OO.

  • EE: Probability is 1/6
  • EO: Probability is 1/3
  • OE: Probability is 1/6
  • OO: Probability is 1/3

The total probability adds up to 1 (1/6 + 1/3 + 1/6 + 1/3 = 1). The first three scenarios (EE, EO, OE) result in an even product xy. Therefore, scenarios 2 and 3 do not include scenario 1.


hi Bunuel
say case 2, x=even, y = whatever
then x is selected from 2 or 4,
y selected from 5,6,7 -- so it is highly possible to picked up x=2, y=6

say case 1, both x and y are even
so it is possible x=2,y=6,

you seem case 1 is already included in case 2

that's why I didn't consider case1,

genuinely need your further clarify.
thanks a lot


The second case is not 'x = even, y = any number'; it is 'x = even AND y = odd'. This means that the probability of this case does not cover the situation where both x and y are even. So, the first case, where both numbers are even, is not included in the second scenario.
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zoezhuyan
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink] New post   02 Dec 2023, 00:51
Bunuel wrote:
zoezhuyan wrote:
Bunuel wrote:
There are four complementary, and thus mutually exclusive, scenarios when choosing two numbers: EE, EO, OE, and OO.

  • EE: Probability is 1/6
  • EO: Probability is 1/3
  • OE: Probability is 1/6
  • OO: Probability is 1/3

The total probability adds up to 1 (1/6 + 1/3 + 1/6 + 1/3 = 1). The first three scenarios (EE, EO, OE) result in an even product xy. Therefore, scenarios 2 and 3 do not include scenario 1.


hi Bunuel
say case 2, x=even, y = whatever
then x is selected from 2 or 4,
y selected from 5,6,7 -- so it is highly possible to picked up x=2, y=6

say case 1, both x and y are even
so it is possible x=2,y=6,

you seem case 1 is already included in case 2

that's why I didn't consider case1,

genuinely need your further clarify.
thanks a lot


The second case is not 'x = even, y = any number'; it is 'x = even AND y = odd'. This means that the probability of this case does not cover the situation where both x and y are even. So, the first case, where both numbers are even, is not included in the second scenario.


hi @Bunue thanks for your help
I have no idea where is wrong about my approach
if x=even, y=any number, then there are 1C2*1C3 = 6 pairs
if y=even, x=any number , then there are 1C4 = 4 pairs
so total 10 pairs.

there are total 1C4*1C3 pairs if pick up x, y randomly

so the possibility = 10 pairs / 12 pairs = 5/6 = E

genuinely want your clarification.

thanks a lot
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Bunuel
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink] New post   02 Dec 2023, 01:58
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zoezhuyan 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


hi @Bunue thanks for your help
I have no idea where is wrong about my approach
if x=even, y=any number, then there are 1C2*1C3 = 6 pairs
if y=even, x=any number , then there are 1C4 = 4 pairs
so total 10 pairs.

there are total 1C4*1C3 pairs if pick up x, y randomly

so the possibility = 10 pairs / 12 pairs = 5/6 = E

genuinely want your clarification.

thanks a lot


The number of pairs with an even number from {1, 2, 3, 4} and any number from {5, 6, 7} is 2*3 = 6:
    (2, 5)
    (2, 6)
    (2, 7)
    (4, 5)
    (4, 6)
    (4, 7)

The number of pairs with any number from {1, 2, 3, 4} and an even number from {5, 6, 7} is 4*1 = 4:
    (1, 6)
    (2, 6)
    (3, 6)
    (4, 6)

From 6 + 4 = 10, we should subtract the overlap when picking an even number from both sets, which is 2*1 = 2 pairs ((2, 6) and (4, 6)). Therefore, the number of pairs giving an even product is 10 - 2 = 8:
    (2, 5)
    (2, 6)
    (2, 7)
    (4, 5)
    (4, 6)
    (4, 7)
    (1, 6)
    (3, 6)

The total number of pairs is 4*3 = 12.

Therefore, the probability that xy will be even is 8/12 = 2/3.

Answer: D.

I hope this explanation has clarified the matter.
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zoezhuyan
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink] New post   03 Dec 2023, 04:25
Bunuel wrote:
zoezhuyan 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


hi @Bunue thanks for your help
I have no idea where is wrong about my approach
if x=even, y=any number, then there are 1C2*1C3 = 6 pairs
if y=even, x=any number , then there are 1C4 = 4 pairs
so total 10 pairs.

there are total 1C4*1C3 pairs if pick up x, y randomly

so the possibility = 10 pairs / 12 pairs = 5/6 = E

genuinely want your clarification.

thanks a lot


The number of pairs with an even number from {1, 2, 3, 4} and any number from {5, 6, 7} is 2*3 = 6:
    (2, 5)
    (2, 6)
    (2, 7)
    (4, 5)
    (4, 6)
    (4, 7)

The number of pairs with any number from {1, 2, 3, 4} and an even number from {5, 6, 7} is 4*1 = 4:
    (1, 6)
    (2, 6)
    (3, 6)
    (4, 6)

From 6 + 4 = 10, we should subtract the overlap when picking an even number from both sets, which is 2*1 = 2 pairs ((2, 6) and (4, 6)). Therefore, the number of pairs giving an even product is 10 - 2 = 8:
    (2, 5)
    (2, 6)
    (2, 7)
    (4, 5)
    (4, 6)
    (4, 7)
    (1, 6)
    (3, 6)

The total number of pairs is 4*3 = 12.

Therefore, the probability that xy will be even is 8/12 = 2/3.

Answer: D.

I hope this explanation has clarified the matter.



opps, got it,

thanks . I neglected the overlap.
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ruis
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink] New post   04 Dec 2023, 14:26
How would this question be solved with Combinatorics?
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink] New post   04 Dec 2023, 14:58
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ruis wrote:
How would this question be solved with Combinatorics?


1 - P(opposite event) = 1 - 2C1/4C1*2C1/3C1 = 1 - 2/4*2/3 = 8/12 = 2/3.
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
04 Dec 2023, 14:58
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