To determine the probability of a randomly chosen 2-digit number being divisible by 3 , we first need to understand the range of 2-digit numbers and then figure out how many of these are divisible by 3.
Let's consider the range of 2-digit numbers; these numbers range from 10 to 99 . Thus, there are:
99−10+1=90two-digit numbers.
Now, we need to determine how many of these numbers are divisible by 3 . A number is divisible by 3 if the sum of its digits is divisible by 3 .
The first 2-digit number divisible by 3 in the range 10 to 99 is 12 . The sequence of 2-digit numbers divisible by 3 then proceeds as
12,15,18,...,99.
This sequence is an arithmetic sequence where the first term,
a, is 12 and the common difference,
d, is 3 . We can use the formula for the
n-th term of an arithmetic sequence:
an=a+(n−1)dwhere
an is the last term. Setting
an=99, we have:
99=12+(n−1)3Solving for
n, we get:
‌99=12+3n−3‌99=9+3n ‌90=3n‌n=30So, there are 30 2-digit numbers that are divisible by 3 .
Therefore, the probability
P of choosing a number that is divisible by 3 is given by the ratio of favorable outcomes to the total number of outcomes:
Hence, the correct option is:
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