A number is randomly chosen in interval 1 to 3. what is prob that first digit to right of decimal is 5

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A number is randomly chosen in interval 1 to 3. what is prob that first digit to right of decimal is 5

My attempt:

1.0, 1.1 ....3.0 --> total there are 21 digits

favourable outcomes = 1.5, 2.5 --> means there are 2 favourable outcomes.

so required probability = 2/21 = 0.095

But in textbook, he assumed it as uniform distribution and solved it(then the ans = .1). When to apply which distribution, how to interpret? Please elaborate

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Since we only care about the number immediately to the right of the decimal place, we can just look at the range $[0,1)$ instead of $[1,3)$. (The choice of including the endpoints is kind of irrelevant since it makes a difference of 1 number when there are infinite numbers in any range. I picked it the way I did so the process for finding the answer is simpler).

Consider the following number in that range:

$$0.2748191\dotsm$$

Notice how we can make a similar looking number that does fit the requirements:

$$0.5748191\dotsm$$

In fact, for any combination of digits after the first digit after the decimal point, you can make 10 numbers by changing the number after the decimal point.

$$0.0748191\dotsm$$ $$0.1748191\dotsm$$ $$0.2748191\dotsm$$ $$0.3748191\dotsm$$ $$0.4748191\dotsm$$ $$0.5748191\dotsm$$ $$0.6748191\dotsm$$ $$0.7748191\dotsm$$ $$0.8748191\dotsm$$ $$0.9748191\dotsm$$

You’ll see that only 1 out of 10 possibilities satisfy the requirement of having a 5 immediately after the decimal point. Therefore, you get the answer $0.1$