If $W$ represents the player’s score on $1$ spin of the wheel, then what is $Pr[W\le1.5]\text{?}$ A wheel is spun with the numbers $1, 2$, and $3$ appearing with equal probability of $1\over 3$ each. If the number $1$ appears, the player gets a score of $1.0$; if the number $2$ appears, the player gets a score of $2.0$, if the number $3$ appears, the player gets a score of $X$, where $X$ is a normal random variable with mean $3$ and standard deviation $1$. 
If $W$ represents the player’s score on $1$ spin of the wheel, then what is $Pr[W\le1.5]\text{?}$
My thought process is that it is the probability of getting the number $1$ + getting number $3$= $${1\over 3}(P(Z<{1.5-3})+1)$$ Is this right?
 A: It seems that your term is right. Only if the number $1$ or $3$ appears $W$ can be smaller or equal to $1.5$. This is right. Then let $U$ be the random variable for the appearing numbers of the wheel. Therefore 
$P(W\leq 1.5)=P(U=1)+P(X\leq 1.5|U=3)$
$=P(U=1)+P(U=3)\cdot P(X\leq 1.5)$
$=\frac13+\frac13\cdot P(X\leq 1.5)$
$=\frac13+\frac13\cdot \Phi\left(\frac{1.5-3}{1}\right)$
$\frac13+\frac13\cdot \Phi\left(-1.5\right)$
This is equivalent to you term. $\Phi(z)$ is the cdf of the standard normal distribution. Consequently
$P(W\leq 1.5)=\frac13+\frac13\cdot 0.06681$
A: Regards User. If I may contribute. Let $N$ be the random variable for the number obtained in 1 wheel spun. You could use this law, from Bayes theorem :
\begin{align*} \small P ( W < w ) &= \small P ( W < w \: | \:N = 1) \cdot P(N=1) + P ( W < w \: | \: N = 2 ) \cdot P(N=2) \\ &\small + \small P ( W < w \: | \: N = 3) \cdot P(N=3)   \end{align*}
For $ P(W < 1.5)$, we have two direct-known values :


*

*$\small P(W < 1.5 \: | \: N =2) = 0$, (since it is certain that the score is $2.0$, when $N=2$), 

*$ \small P(W < 1.5 \: |  \: N = 1) = 1$
So :
\begin{align*} \small P(W < 1.5) &= \small 1 \cdot \frac{1}{3} + 0 + P(W < 1.5 \: | N = 3) \cdot \frac{1}{3} \\ 
&= \small \frac{1}{3} \left[ 1 + P(W < 1.5 | N =3) \right] 
\end{align*}
$P(W < 1.5 | N =3)$ is as same as $P(X<1.5)$. Now you should be confident to continue from here.
Hope this helps. Regards, Arief.
