Finding uniform distribution beyond a and b values $X$ is uniformly distributed over a range of values from $8$ to $21$. Probabilty of $(X<22)$ and Probability of $(X>7)$? How do i solve when the probabilities are beyond the $a$ and $b$ values?
 A: They are both equal to one.  The uniform distribution is only defined between $8$ and $21$, and is equal to $0$ everywhere else.  So, probability of $X<22$ and $X>7$ is $1$.
The probability of the variable taking on a value outside of the bounds is $0$, so you just ignore it.
A: $P(X < 22) = 1$ as $X$ is uniformly distributed over $8$ to $21$ that is $P(X < 22) = P(8 \leq X \leq 21) = 1$,as $P(X = k)$ where $k \notin [8,21]$ is zero as our $X$ is only distributed over $[8,21]$.Probability of $X > 7$ is same as  $P(8 \leq X \leq 21 )  =1$,that is there is full chance of $X$ having values between $[8,21]$
A: *

*When $X\in [8,21], a\geq 8 \land b \leq 21,$ then $P(a\leq X\leq b) = \frac{b-a}{21-8}$

*When $X\in [8,21],$ for arbitary $ a,$ then $P(a\leq X) = \begin{cases}1 \quad a\leq 8\\ \frac{21-a}{21-8} \quad 8 < a\leq 21 \\ 0 \quad a> 21\end{cases}$

*When $X\in [8,21],$ for arbitary $ b,$ then $P( X\leq b) = \begin{cases}1 \quad 21\leq b\\ \frac{b-8}{21-8} \quad 8 < b\leq 21 \\ 0 \quad 8> b\end{cases}$

A: This has nothing to do with the fact that $X$ is uniformly distributed.
You've been told that $X$ is always between $8$ and $21$, so we can say for sure that $X>7$, thus the probability is $1$. Same goes for $x<22$.
