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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.
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.
- 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.
02 Dec 2023, 00:51
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Bunuel
zoezhuyan
Bunuel
There are four complementary, and thus mutually exclusive, scenarios when choosing two numbers: EE, EO, OE, and OO.
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.
- 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
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
(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.
