Probability in uniform

Question

I need help to solve the following problems. Thank you in advance.

Problem 1:

A random variable $X$ is uniform $[0, 1]$. Find the probability that X's 2nd digit is $3$.

As far as I understand it is continuous uniform distribution. Each digit has $1$ chance in $10$ of being a $3$. Does it mean that the probability of X’s second digit being 3 is $1/10$? Can you check and confirm it?

Problem 2:

A random variable $X$ is uniform $[0, 3]$. Find the probability that X's first and/or second digit is $2$.

As I understand, the first digit has $1$ chance in $3$ of being a $2$, 2nd digit $1$ chance in $10$ of being a $2$. Avoiding both has probability $(2/3) (9/10)$. Contrary event $1− (2/3) (9/10)$. Am I correct? If so, eventually what is the probability that X's first and/or second digit is $2$. Maybe I am dumb and don’t get it, can someone help me?

Answer 1

For problem 1, the only way that the second digit of $X$ can be equal to $3$ is if $X$ belongs to one of the following intervals: $[.03,.04),[.13,.14),[.23,.24),[.33,.34),\ldots,[.93,.94)$. The probability of this happening is the sum of the lengths of these intervals, which is $.01 \times 10 = .1$. This confirms your intuition that the probability is $1/10$.

Answer 2

For problem 2, find the intervals of success and divide by the total interval.

The intervals of success (i.e. the first or the second digit is $2$) are: $$[0.2,0.3); [1.2,1.3); [2,3) \Rightarrow I_S=0.1+0.1+1=1.2$$

The total interval is: $$[0,3] \Rightarrow I_T=3.$$ Hence: $$P(D_1=2\cup D_2=2)=\frac{I_S}{I_T}=\frac{1.2}{3}=\frac25.$$ Note that it is consistent with your answer $1-\frac23\cdot \frac{9}{10}$.