stephny wrote:
if a three-digit number is selected at random from the integers 100 to 999, inclusive, what is the probability that the first digit and the last digit of the integer will be both be exactly two less than the middle digit?
P(both the first digit and the last digit are exactly two less than the middle digit) = (# of integers that meet the given condition)/(total number of integers)
# of integers that meet the given condition
Take the task of creating 3-digit numbers and break it into stages.
Stage 1: Select the hundreds digit
Since the hundreds digit must be 2 less than the tens digit, the hundreds digit can be 1, 2, 3, 4, 5, 6 or 7
So, we can complete stage 1 in
7 ways
Stage 2: Select the hundreds digit
Once the hundreds digit has been selected in stage 1, there's ONLY 1 way to select the tens digit, since the tens digit must be 2 GREATER than the hundreds digit. For example, if the hundreds digit is 6, the tens digit must be 8.
So, we can complete stage 2 in
1 way
Stage 3: Select the units digit
Once the hundreds digit has been selected in stage 1, there's ONLY 1 way to select the units digit, since the units digit must be IDENTICAL to the hundreds digit. For example, if the hundreds digit is 7, the tens digit must be 7 as well.
So, we can complete stage 3 in
1 way
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus build a 3-digit number) in
(7)(1)(1) ways (=
7 ways)
Total number of integers from 100 to 999 inclusive
Rule:
the number of integers from x to y inclusive equals y - x + 1
So, number of integers from 100 to 999 inclusive = 999 - 100 + 1 =
900
P(first digit and the last digit of the integer will be both be exactly two less than the middle digit) = (# of integers that meet the given condition)/(total number of integers)
= [spoiler]
7/
900[/spoiler]
Cheers,
Brent
Aside: For more information about the FCP, watch our free video:
https://www.gmatprepnow.com/module/gmat-counting?id=775
