total integers ; 250-101+1 = 150
with tens digit as 1 would be 20 such digits
P ; 20/150 = 2/15 IMO B
with tens digit as 1 would be 20 such digits
P ; 20/150 = 2/15 IMO B
energetics wrote:
What is the probability that a random number selected from the set of integers 101 through 250, inclusive, has a tens digit of 1?
(A) \(\frac{1}{15}\)
(B) \(\frac{2}{15}\)
(C) \(\frac{1}{25}\)
(D) \(\frac{3}{25}\)
(E) \(\frac{20}{149}\)
(A) \(\frac{1}{15}\)
(B) \(\frac{2}{15}\)
(C) \(\frac{1}{25}\)
(D) \(\frac{3}{25}\)
(E) \(\frac{20}{149}\)
Spoiler: ::
EXPL. First, determine the total number of integers between 101 and 250, inclusive. The difference between 250 and 101 is 149, but because the endpoints are included, the total number of integers is one greater than that: 150.
Second, figure out how many of those 150 integers have a tens digit of 1. In the 100s (100 to 199, inclusive), there are 10 numbers with a tens digit of 1: the integers between 110 and 119, inclusive. The same goes for the 200s: we can include all the integers between 210 and 219, inclusive. That gives us a total of 20 desired outcomes.
The probability, then, is desired over possible: p = 20/150 = 2/15 , choice (B).
Second, figure out how many of those 150 integers have a tens digit of 1. In the 100s (100 to 199, inclusive), there are 10 numbers with a tens digit of 1: the integers between 110 and 119, inclusive. The same goes for the 200s: we can include all the integers between 210 and 219, inclusive. That gives us a total of 20 desired outcomes.
The probability, then, is desired over possible: p = 20/150 = 2/15 , choice (B).