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30 Jun 2025, 15:10
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Let N be the total number of integers from x to y, inclusive.
The number of integers from x to y, inclusive, is given by N=y−x+1.
(1) y - x = 9
From this, we can find N:
N=y−x+1=9+1=10.
we can calculate the probability
Since we can find a unique probability, statement (1) is sufficient.
(2) y = -20
This statement only gives the value of y. We still don't know x, and therefore we don't know N.
Since we get different probabilities depending on x, statement (2) is not sufficient.
The number of integers from x to y, inclusive, is given by N=y−x+1.
(1) y - x = 9
From this, we can find N:
N=y−x+1=9+1=10.
we can calculate the probability
Since we can find a unique probability, statement (1) is sufficient.
(2) y = -20
This statement only gives the value of y. We still don't know x, and therefore we don't know N.
Since we get different probabilities depending on x, statement (2) is not sufficient.
30 Jun 2025, 15:19
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Statement 1 alone is sufficient , but statement 2 alone is not.
1- Alone we are able to plug in many number optiosn all resulting in y>x. y=100 x=91 , y=-100 x=-109 y=10, x=1, y=11, x=2, y=-12 x=-21
2- We are unable to determine what x is and cant come to a conclusion for y > or < x.
1- Alone we are able to plug in many number optiosn all resulting in y>x. y=100 x=91 , y=-100 x=-109 y=10, x=1, y=11, x=2, y=-12 x=-21
2- We are unable to determine what x is and cant come to a conclusion for y > or < x.
Bunuel
30 Jun 2025, 15:54
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Ann and Bob each randomly select an integer from the integers x to y, inclusive. What is the probability that the integer Ann selects is greater than the one Bob selects?
numbers n=y-x+1
n^2 is the total number os odds.
There are exactly n odds that Ann and Bobb choose the same number. In the rest of odds, there is the same probability that Ann chooses a greater number that the one Bob chooses as the opposite happening, so:
P = (n^2-n)/2*n^2 = (n-1)/2n
1) y-x=9
n=10
P=9/20
SUFFICIENT
2)y=-20
No way to find n
INSUFFICIENT
IMO A
numbers n=y-x+1
n^2 is the total number os odds.
There are exactly n odds that Ann and Bobb choose the same number. In the rest of odds, there is the same probability that Ann chooses a greater number that the one Bob chooses as the opposite happening, so:
P = (n^2-n)/2*n^2 = (n-1)/2n
1) y-x=9
n=10
P=9/20
SUFFICIENT
2)y=-20
No way to find n
INSUFFICIENT
IMO A
