Mathematical Statistics (Wackerly, Mendenhall, Scheaffer): Problem 4.98: Mixed Probability Distribution Problem Statement. The duration $Y$ of long-distance telephone calls (in minutes) monitored by a station
is a random variable with the properties that
$$P(Y=3)=0.2\qquad\text{and}\qquad P(Y=6)=0.1.$$
Otherwise, $Y$ has a continuous density function given by
$$
f(y)=
\begin{cases}
(1/4)ye^{-y/2},&y>0\\
0,&\text{elsewhere.}
\end{cases}
$$
The discrete points at $3$ and $6$ are due to the fact that the length of call is announced to the
caller in $3$-minute intervals and the caller must pay for $3$ minutes even if he or she talks
less than $3$ minutes. Find the expected duration of a randomly selected long-distance call.
My Thoughts So Far. The incredibly annoying thing about this problem is that there is overlap between the discrete distribution given at $3$ and $6,$ and the continuous distribution given for all positive $Y.$ The problem statement appears to be ambiguous. Moreover, if you integrate $f(y)$ from $0$ to $\infty,$ you find that it is $1.$ So $f(y)$ is a correctly given density function in its own right. Somehow we have to balance these two. My initial thought is to do the following. 1. Create a new discrete distribution as follows:
$$P(Y=3)=\frac23\qquad\text{and}\qquad P(Y=6)=\frac13.$$
We can call this $g(y),$ so that we can balance an synthesized function as
$$h(y)=(0.3)g(y)+(0.7)f(y).$$
Here I got the $0.3$ as $0.2+0.1$ in the original discrete distribution.

Is this the correct approach? It basically ignores the overlap and simply allows it. Is that a problem?

Thank you for your time!
 A: The "real" way to think of this is in terms of what probability we're assigning to certain sets; formally you could do this with measure theory and use a dominating measure that's a mixture of point masses at $3$ and $6$ plus the Lebesgue measure on $\mathbb R$, but we don't need that formality to reason about this correctly and get the answer.
We know that the set $[0,\infty)$ needs a probability of $1$, so 
$$
1 = P(Y \in [0,\infty)) = P(Y = 3) + P(Y = 6) + P(Y \in [0,3)\cup(3,6)\cup(6,\infty))
$$
since those are disjoint events. The continuous density has anything like $P(Y=y)$ being zero, so $P(Y=3)$ and $P(Y=6)$ are purely determined by the discrete point masses that you're given. That's a key point here and is why the overlap doesn't really matter. Thus
$$
1 = P(Y \in [0,\infty)) = 0.3 + P(Y \in [0,3)\cup(3,6)\cup(6,\infty))
$$
so
$$
P(Y \in [0,3)\cup(3,6)\cup(6,\infty)) \\
= P(Y \in [0,3)) + P(Y \in (3,6)) + P(Y > 6) \\
= 0.7
$$
and on these sets the probability is purely governed by the continuous density. 
This lets us work out the CDF of $Y$:
$$
F(y) = P(Y \leq y) = 0.7 \int_0^y \frac 14 te^{-t^2/2}\,\text dt + 0.2 \cdot \mathbf 1_{y \geq 3} + 0.1 \cdot \mathbf 1_{y \geq 6}.
$$
This smoothly increases up to $y=0.2$ when it jumps by $0.2$. There's no issue for the continuous density because the set $\{3\}$ has probability zero w.r.t. that density. Then it smoothly increases up to $y=6$ where there's again a jump, this time of $0.1$. 
To finish off the problem, you can directly get the expected value from a CDF or we can just get a density that works.
Again, without measure theory we can't be really formal about what we're doing here, but we can "guess" the correct density of $Y$ from this because $F$ is differentiable almost everywhere. When $y \notin \{3,6\}$ $F$ is differentiable, and via the fundamental theorem of calculus we have
$$
f(y) = F'(y) = 0.7 \cdot \frac 14 ye^{-y^2/2}
$$
(for $y \geq 0$). This is "mixed" with the point masses at $3$ and $6$, so we can integrate $y$ times the density (I'm being vague about what we're integrating with respect to) to get
$$
\text E(Y) = \int y f(y)\, "\text dy" \\
= 0.7 \int_{y \notin \{3,6\}}y \cdot \frac 14 ye^{-y^2/2}\,\text dy + 3 \cdot 0.2 + 6 \cdot 0.1 \\
= \frac{0.7}4 \int_0^\infty y^2e^{-y^2/2}\,\text dy + 1.2 
$$
and that is just a definite integral at this point.
