Question about the support of two random variables

Question

Here's what the question asked:

"Suppose that a bowl contains 100 chips: 30 are labelled 1, 20 are labelled 2, and 50 are labelled 3. The chips are thoroughly mixed, a chip is drawn, and the number X on the chip is noted.

(c) Compute $ P(W = w)$ for every real number $w$ when $W = X + Y$."

Here's my work:

$X$ represents the number on the chip, and so does $Y$. So the support of $X$, $\Large\chi$, is the following: $$\chi \in \{1,2,3\}$$

Similarly, the support of Y, the number on the second chip, $\Large\gamma$, is the following: $$\gamma \in \{1,2,3\}$$ So I was thinking that the sample space of $W$ could be the following: $$\{2,3,4,3,4,5,4,5,6\}$$

After removing the duplicates, the support of W, $\Large\Omega$, could be: $$\{2,3,4,5,6\}$$

Thus: $$Pr(\Omega = \omega) = \left\{ \begin{array}{rcl} \frac{\omega-1}{9} & \mbox{if} & 3\leq\omega\leq4 \\ \frac{(\omega-1)-2^{2^{(\omega-5)}}}{9} & \mbox{if} & 4<\omega\leq6 \\ 0 && \mbox{otherwise} \end{array}\right.$$

In other words: $$Pr(\Omega=2)=\frac{1}{9} \\ Pr(\Omega=3)=\frac{2}{9}\\Pr(\Omega=4)=\frac{3}{9} \\Pr(\Omega=5)=\frac{2}{9} \\Pr(\Omega=6)=\frac{1}{9}$$

The book's answers were different: $$Pr(\Omega=2)=.09\\Pr(\Omega=3)=.12\\Pr(\Omega=4)=.34\\Pr(\Omega=5)=.2\\Pr(\Omega=6)=.25$$

What am I doing wrong?

Answer 1

In order to get the answer in the book, the answer to the question in my Comment has to be two draws with replacement.

First, a simulation of a million draws of two chips with replacement, which should give answers correct to two or three places.

chips = rep(1:3, c(30,20,50))
set.seed(922)
m = 10^6
w = replicate(m, sum(sample(chips,2, repl=T)))
table(w)/m
w
       2        3        4        5        6 
0.089752 0.119838 0.339444 0.200383 0.250583 

Which round to $.09, .12, .34, .20,$ and $.25,$ respectively.

By combinatorics:

• How to get event $\{W=2\}?$ That takes two 1's: $(.3)^2 = .09.$

• How to get event $\{W=3\}?$ That takes 12 or 21: $2(.3)(.2) = .12.$

• How to get event $\{W=4\}?$ That takes 13, 31 or 22: $2(.3)(.5)+(.2)^2 = .34.$

And so on for getting a total of 5 or 6.

Note: Answers are a little different if sampling is without replacement (which is the default mode of sample in R:

chips = rep(1:3, c(30,20,50))
set.seed(2020)
m = 10^6
w = replicate(m, sum(sample(chips,2)))
table(w)/m
w
       2        3        4        5        6 
0.087738 0.121141 0.341016 0.202335 0.247770 

Rounding to two places still gives the same answers, but the precise answers are a little different.

For example: $P(W = 2) = (30/100)(29/99) = 0.08787879.$