The random variable U is uniformly distributed over the interval [3,6]. Find the following probabilities

Question

(a) $P[\frac4 5 ≤ U ≤ 4]$ = $$P[\frac4 5 ≤ U ≤ 3]+P[3≤ U ≤ 4]=0+P[U ≤ 4]=\frac{(4-3)}{(6-3)}=\frac1 3$$ Is this even right? I'm just not so confident.

(b) $P[U > 5]$ = $$1-P[U ≤5]=1-\frac{8-3}{6-3}=1-\frac5 3$$ The answer for b is a positive number according to my friends so I am not sure that I'm correct in this one either.

(c) $P[16 ≤ U^2 ≤ 36]$=

First I simplify them

$$P[4≤ U ≤ 6] or P[-6 ≤ U ≤ -4]$$

Then I form the equation

$$P[16 ≤ U^2 ≤ 36]= P[4≤ U ≤ 6]+P[-6 ≤ U ≤ -4]$$ But I am not sure what to do afterwards, do I have to integrate it? If so, how??

(d) $P[4−2|U|≥−8$= $$ P[|U|≤ 5] = P[−5 ≤ U ≤ 5] = P[3 ≤ U ≤ 5] = 1/3$$ Is this even correct??

Trying to do the exercises based on my friend's notes on the classes. So far I still feel lost on these kinds of exercises but the answers just seem so off so far so I would appreciate some advice.

Answer 1

$(a)$ is correct.
$(b)$ $$1-P[U ≤5]=1-\frac{\textbf{5}-3}{6-3}=\frac1 3$$
$(c)$ $$P[16 ≤ U^2 ≤ 36]= P[4≤ U ≤ 6]+P[-6 ≤ U ≤ -4] = P[4≤ U ≤ 6]+0 = 1-P[U ≤4]=1 - \frac{4-3}{6-3} = \frac2 3$$
$(d)$ $$P[4−2|U|≥−8] = P[-2|U|≥-12]=P[-|U|≥-6] =P[|U|≤6]=P[U≤6]= 1 $$

Answer 2

Note that your probability density function is $f(x)= 1/3$ for $x\in[3,6]$ and $f(x)=0$, otherwise.

$(a)$ $$P[\frac4 5 ≤ U ≤ 4]= P[ 3 ≤ U ≤ 4]=1/3$$

$(b)$ $$P[U > 5]=P[5<U<6]=1/3$$

$(c)$ $$P[16 ≤ U^2 ≤ 36]= \\P[4≤ U ≤ 6]+P[-6 ≤ U ≤ -4]=\\2/3+0=2/3$$