Information given:- Ann & Bob each randomly pick an integer from x to y, inclusive
- We want the probability that Ann's number > Bob's number
- Statement 1: y - x = 9
- Statement 2: y = -20
Question:- Is the probability that Ann's number is greater than Bob's number (uniquely) determinable?
Solution:- The probability depends on the range size, e.g. the number of integers possible
- Total possible outcomes: n^2, where n = y - x + 1 (this is the number of 'options' that Ann and Bob have)
- Therefore, to find the probability that Ann's number > Bob's number, you need n
- Statement 1: y - x = 9
- n = (y - x) + 1 = 9 + 1 = 10
- We know how many integers there are, so we can calculate the probability that Ann's number > Bob's number
- P (Ann > Bob) = favorable pairs / (10^2)
- Therefore, statement 1 is sufficient
- Statement 2: y = -20
- However, x is unknown, so we can't find the range or n (see explanation under statement 1)
- For example, if x = -29, then n = 10
- If x = -50, then n is different
- The probability that Ann picks a greater number than Bob can change
- Therefore, statement 2 is insufficient
- Since statement 1 alone is sufficient, statement 1 and 2 together are also sufficient
Answer: A, statement 1 alone is sufficientBunuel
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?
(1) y - x = 9
(2) y = -20