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555-605 (Medium)|   Probability|                                    
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Bunuel
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 30 Nov 2023, 05:38
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zoezhuyan
Bunuel

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
Show more

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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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 02 Dec 2023, 00:51
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Bunuel
zoezhuyan
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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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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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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 02 Dec 2023, 01:58
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zoezhuyan
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
Show more

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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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 04 Dec 2023, 14:26
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How would this question be solved with Combinatorics?
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 04 Dec 2023, 14:58
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ruis
How would this question be solved with Combinatorics?
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1 - P(opposite event) = 1 - 2C1/4C1*2C1/3C1 = 1 - 2/4*2/3 = 8/12 = 2/3.
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 13 Sep 2024, 04:30
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Abdul29
x*y is even only when they're both even or one of them is odd. The probability of any single outcome is given by 1/3*1/4 = 1/12.
Working through possible outcomes, we arrive at 8/12 -> 2/3, hence (D).

It took me around a minute and 40 seconds to solve this, I'm sure that a faster approach exists, waiting for others.
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Yup, I got it through the same method, 1/4 is the probability for any number on the 1st set, then I check the probability of getting an even by multiplying each number in the first set to each one in the second. It includes a lot of steps but it is straightforward.
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 13 Oct 2024, 02:59
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Two cases:

1. Odd from first - even from second
2. Even from first - any from second

2C1. * 1C1 + 2C1*3C1

Total = 4C1*3C1

Divide and you will get the answer
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 07 Jul 2025, 08:57
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To find the probability that x × y is even, where x is chosen from {1, 2, 3, 4} and y from {5, 6, 7}:

1. Total combinations: 4 × 3 = 12.
2. Odd combinations (where x × y is odd):
- Odd x: 1, 3 (2 choices)
- Odd y: 5, 7 (2 choices)
- Total odd combinations: 2 × 2 = 4.
3. Even combinations: 12 - 4 = 8.
4. Probability that x × y is even:
\( \frac{8}{12} = \frac{2}{3} \)

Final answer: \(\frac{2}{3} \)
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 08 Jul 2025, 00:05
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Another intuitive way to solve this problem is...

Case 1: x even * y odd or even.
- P(x even)*P(y odd or even) = 2/4 * 1 = 1/2

Case 2: x odd or even * y even.
- P(x odd or even)*P(y even) = 1 * 1/3 = 1/3

Intersection: x even * y even.
- P(x even)*P(y even): 1/2 * 1/3 = 1/6

So... Case 1 + Case 2 - Intersection = 1/2 + 1/3 - 1/6 = 3/6 + 2/6 - 1/6 = 4/6 = 2/3.
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 12 Jul 2025, 08:22
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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?

For XY to be even, there are two category of possibilities:

A) Even from 1st set (2,4) multiplied by any number (5,6,7) from second set (Since even by odd/even gives even)
B) Odd from 1st set (1,3) multiplied by even (6) from second set (Since odd*even gives even, HERE we are not considering 6*2 and 6*4 since they are considered in Category A

P(A)= (2/4)*(3/3)
P(B)=(2/4)*(1/3)
P(Even)= P(A)+P(B)= 2/3
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If x is to be chosen at random from the set {1, 2, 3, 4} and y is to 12 Jul 2025, 12:07
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I have two cases, based on the outcome of set 1:

1) What I pick from set 1 is odd: 1/2 -> the only way I have an even product will be choosing an even number on set 2 (probability of this = 1/3) = total probability = 1/6

2) What I pick from set 1 is even: 1/2 -> I will always have an even product, no matter what number I choose on the second set -> total probability is 1/2 x 1 = 1/2

I add the two probabilities to have a total of 1/6 + 1/2 = 2/3

Answer (D).
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