Let $X$ be a random variable which is uniformly distributed on the interval $[0, 1]$. Let $Y$ be the first digit of $X$. Show that $P(Y = 0) = .1$ (Hint: $P(Y = 0) = P(0\le X <.1)$ Why?).
Hello, I don't really understand how to approach this question here. Uniform distribution means that the probability of anything is $1/(b - a)$ and in this case $b = 1$ and $a = 0$. But What do they mean by "first digit of $X$"? And why is $ P(Y = 0) = P(0<=X <.1)$, or specifically, why is $.1$ used and not $0.01$ or even $0.001$?
The question is slightly unclear.
By the first digit, it means the first digit after the decimal point.
Example of numbers between $[0,0.1)$ and numbers such as $0.\color{blue}01$ and $0.\color{blue}03$.
Example of numbers between $[0.1, 0.2)$ numbers such as $0.\color{blue}12$ and $0.\color{blue}15$.
With that hopefully you can answer the question.
Extra question: What is the probability that the first digit begins with $5$?