I am working through an old college book (Probability & Statistics for Engineering and the Sciences). I came across the following problem and gave myself the extra challenge of creating the entire joint pmf. My problem is that my numbers are not adding up correctly even though I'm getting the correct answers to the given question. I'll write the problem and then explain my work. (Note: I am using $n$C$k$ to denote $n$ choose $k$ since I can't figure out how to do the choose notation in mathjax.)
Problem: The number of customers waiting for gift-wrap service at a department store is an rv $X$ with possible values $0, 1, 2, 3, 4$ and corresponding probabilities $.1,.2,.3,.25,.15$. A randomly selected customer will have $1,2,$ or $3$ packages for wrapping with probabilities $.6, .3,$ and $.1$ respectively. Let $Y =$ the total number of packages to be wrapped for the customers waiting in line (assuming that the number of packages submitted by one customer is independent of number submitted by any other customer).
a. Determine $P(X = 3,\ Y = 3)$, i.e. $p(3,3).$
b. Determine $p(4,11)$.
c. (I'm doing this part for my own sake) Create the joint pmf of $X$ and $Y$.
Solution:
a. For $p(3,3)$ there are $3$ customers with one gift each. The probability of this event happening is $0.25\times0.6^3= 0.054$, which is the correct answer.
b. For $p(4,11)$ there must be $3$ customers with $3$ packages each and one customer with two packages. The probability of this happening is $0.15\times 4$C$1 \times .1^3 \times .3 = 0.00018 $, which is the correct answer.
ACTUAL QUESTION
c. This is the PMF as I currently know it to be correct (with $X$ as the rows and $Y$ as the columns):
$$\begin{array}{| c | c | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline
0 & 0.1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline
1 & 0 & .12 & .06 & .02 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline
2 & 0 & 0 & & & & & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline
3 & 0 & 0 & 0 & 0.054 & & & & & & & 0 & 0 & 0\\ \hline
4 & 0 & 0 & 0 & 0 & & & & & & & & 0.00018 & \\ \hline
\end{array}$$
I know $p(0,0), p(3,3)$ and $p(4,11)$ because they were given to me. For $X = 1$ if calculated the probabilities in the following way based on the correct answers to a. and b.:
$$\begin{align*}
p(1,1) &= 0.2 * 0.6 = 0.12\\
p(1,2) &= 0.2 * 0.3 = 0.06\\
p(1,3) &= 0.2 * 0.1 = 0.08
\end{align*}$$
Since these add up to $0.2$, I know that I have the correct probability distribution corresponding to the row for $X = 1$.
For the row $X = 2$, I use the same pattern I used in parts a. and b. but do not come up with a correct distribution. Instead, I get the following:
$$\begin{align*}
p(2,2) &= 0.3 * 0.6^2 = 0.108\\
p(2,3) &= 0.3 * 2\text{C}1 * 0.6 * 0.3 = 0.108\\
p(2,4) &= 0.3 * 0.3^2= 0.027\\
p(2,5) &= 0.3 * 2\text{C}1 * 0.3 * 0.1 = 0.018\\
p(2,6) &= 0.3 * .1^2 = 0.003
\end{align*}$$
The proabilities for $P(X = 2)$ add up to $0.264$. Since this does not add up to $0.3$ I know that I did something incorrectly.
However, I have no idea what it is. I used the same format and counting rules used in parts a. and b., which the book gives as correct answers. What am I missing in my understanding of probability?
Best Answer
For part c, when calculating $p(2,4)$, you forgot to include the possibility that one customer has 3 packages and the other 1 package. This term is $0.3 * 2C1 * 0.1 * 0.6=0.036$.