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30 Jun 2025, 10:16
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Looking at this problem, I need to find the probability that Ann's randomly selected integer is greater than Bob's.
Let me think about this systematically. If there are n integers from x to y inclusive, then n = y - x + 1.
For any two randomly selected integers from the same range, there are three possible outcomes:
Ann's number > Bob's number
Ann's number = Bob's number
Ann's number < Bob's number
By symmetry, P(Ann > Bob) = P(Bob > Ann).
Since all probabilities sum to 1:
P(Ann > Bob) + P(Ann = Bob) + P(Ann < Bob) = 1
2 × P(Ann > Bob) + P(Ann = Bob) = 1
So P(Ann > Bob) = [1 - P(Ann = Bob)]/2
Now I need P(Ann = Bob). There are n ways for them to pick the same number (one for each integer in the range), and n2 total possible outcomes.
So P(Ann = Bob) = n/n2 = 1/n
Therefore: P(Ann > Bob) = [1 - 1/n]/2 = (n-1)/(2n)
Now let me check each statement:
(1) y - x = 9
This tells me there are n = 9 + 1 = 10 integers in the range.
P(Ann > Bob) = (10-1)/(2×10) = 9/20
This is sufficient to answer the question.
(2) y = -20
This only gives me the upper bound. Without knowing x, I can't determine how many integers are in the range, so I can't calculate the probability.
This is not sufficient.
Let me think about this systematically. If there are n integers from x to y inclusive, then n = y - x + 1.
For any two randomly selected integers from the same range, there are three possible outcomes:
Ann's number > Bob's number
Ann's number = Bob's number
Ann's number < Bob's number
By symmetry, P(Ann > Bob) = P(Bob > Ann).
Since all probabilities sum to 1:
P(Ann > Bob) + P(Ann = Bob) + P(Ann < Bob) = 1
2 × P(Ann > Bob) + P(Ann = Bob) = 1
So P(Ann > Bob) = [1 - P(Ann = Bob)]/2
Now I need P(Ann = Bob). There are n ways for them to pick the same number (one for each integer in the range), and n2 total possible outcomes.
So P(Ann = Bob) = n/n2 = 1/n
Therefore: P(Ann > Bob) = [1 - 1/n]/2 = (n-1)/(2n)
Now let me check each statement:
(1) y - x = 9
This tells me there are n = 9 + 1 = 10 integers in the range.
P(Ann > Bob) = (10-1)/(2×10) = 9/20
This is sufficient to answer the question.
(2) y = -20
This only gives me the upper bound. Without knowing x, I can't determine how many integers are in the range, so I can't calculate the probability.
This is not sufficient.
30 Jun 2025, 10:23
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I picked A) Statement 1 Alone is sufficient but statement 2 is not.
With the information given in the stimulus, I wrote the following before looking at the two statements:
1) y - x + 1 = total number of integers in the set x - y, inclusive (?)
2) A>B?
Statement 1:
y - x = 9 can be transformed back into the formula for total number of elements in a set, inclusive of both endpoints: y - x + 1 = 10.
Since we know there are 10 numbers in the set we know exactly half of the set is smaller than the other half of the set.
Statement 2:
y = -20 does not provide enough information to solve for the total number of elements in the set:
-20 - x + 1 = ?
-19 - x = ?
We still need to know something about the endpoint/variable x to solve for the total elements.
With the information given in the stimulus, I wrote the following before looking at the two statements:
1) y - x + 1 = total number of integers in the set x - y, inclusive (?)
2) A>B?
Statement 1:
y - x = 9 can be transformed back into the formula for total number of elements in a set, inclusive of both endpoints: y - x + 1 = 10.
Since we know there are 10 numbers in the set we know exactly half of the set is smaller than the other half of the set.
Statement 2:
y = -20 does not provide enough information to solve for the total number of elements in the set:
-20 - x + 1 = ?
-19 - x = ?
We still need to know something about the endpoint/variable x to solve for the total elements.
30 Jun 2025, 10:28
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From statement (1) we know that there are 10 integers from x to x+9. We can create a table where:
The number of satisfied cases start with x+1, then increases one each row until x+9. This is basically the sum of first 9 natural numbers = \(\frac{(9)*(10)}{2}\).
Ex: y-x=4 so y=x+4
Satisfied (S) = 0+1+2+3+4 = (4*5)/2 = 10
Answer: A
- 1st column represents the numbers Ann can choose.
- 1st row represents the numbers Bob can choose.
The number of satisfied cases start with x+1, then increases one each row until x+9. This is basically the sum of first 9 natural numbers = \(\frac{(9)*(10)}{2}\).
Ex: y-x=4 so y=x+4
Satisfied (S) = 0+1+2+3+4 = (4*5)/2 = 10
| x | x+1 | x+2 | x+3 | x+4 | |
| x | |||||
| x+1 | S | ||||
| x+2 | S | S | |||
| x+3 | S | S | S | ||
| x+4 | S | S | S | S |
Answer: A
30 Jun 2025, 10:31
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Total number in set = y-x +1 (since x and y are inclusive)
Lets say y-x+1=n
Ann select a, Bob selects b. Probability of a>b ?
total no of outcomes = (y-x+1)(y-x+1)= \(n^2\)
no of time a=b will equal to n= y-x+1 (total numbers in set, each pair when both equal)
remaining pairs => \(n^2\)-n
half of pair will have a>b and half will have b>a.
P(a>b) = (\(n^2\) - n)/2*\(n^2\)..... Q
A. y-x = 9 => n=10 we can calculate Q. SUFFICIENT
B. y = -20, to calculate n we need x. NOT SUFFICIENT
Answer A
Lets say y-x+1=n
Ann select a, Bob selects b. Probability of a>b ?
total no of outcomes = (y-x+1)(y-x+1)= \(n^2\)
no of time a=b will equal to n= y-x+1 (total numbers in set, each pair when both equal)
remaining pairs => \(n^2\)-n
half of pair will have a>b and half will have b>a.
P(a>b) = (\(n^2\) - n)/2*\(n^2\)..... Q
A. y-x = 9 => n=10 we can calculate Q. SUFFICIENT
B. y = -20, to calculate n we need x. NOT SUFFICIENT
Answer A
Bunuel
