A and B, both have to select any integer from x to y, inclusive
The total number of integers is given by n = y-x+1 > we add 1 to account for the last digit since both x and y are included in this range
The total number of selections that can be made = (Total # of choices that A has)*(Total # of choices that B has) = \(n^2\)
The probability that A's choice is greater than B's choice is given by:
\(\frac{number of favorable outcomes}{n^2}\)
Let's look at the statements now:
Statement 1:
y−x=9
This gives us, n=y-x+1 = 9+1 = 10
Now, that we have 10 (meaning their are 10 integers for both A and B to select from), let's calculate the probability:
Assume, x=0 and y=10
- If B picks 10, 0 favorabe outcome (since, A cannot pick an integer greater than 10)
- If B picks 9, 1 favorable outcome (since A can pick 10)
- If B picks 8, 2 favorable outcomes (since A can pick any of 9 or 10)
- And so on..
Adding up all the favorable outcomes we get = 0+1+2+....9
This is nothing but the sum of N natural numbers given by \(\frac{N*(N+1)}{2}\)
Since N=9, Sum=\(\frac{9*10}{2}\)=45
The probability is now: \(\frac{number of favorable outcomes}{n^2}\) = \(\frac{45}{100}\) = 0.45
Sufficient.
Statement 2:
y=−20
This statement tells us nothing about x, and consequently, nothing about the total number of integers we have to choose from. Hence, insufficient.
Correct answer is A.