Understanding how to compute $P(W = w)$, given that $W = X + Y$

Question

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. Also, assume that the first chip is replaced, a second chip is drawn, and the number $Y$ on the chip noted.

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

I am really confused as to what $P(W=w)$ means. For example, if we considered $w = 1$, could we write $P(W=1)$ as $P(X = 1) + P(Y = 1)$?

At the back of the textbook where the solutions are written up for that question, it says that $P(W=2) = 0.09$, but doesn't mention anything about $P(W=1)$. Why is that, and how can I go about calculating $P(W=w)$? I know how to calculate $P(X=x)$ and $P(Y=y)$, as we're dealing with one random variable, but when it comes to two, I am just lost.

Answer 1

We want to compute $P(W=w)$ where $W=X+Y$ and $P(W=w)$ is the probability of the event that the random variable $W $ is equal to the value $w$.

Let's first observe that the range of $W $ is the set $\{2,3,4,5,6\}$ since $X,Y$ take values in $\{1,2,3\} $. In particular, $P(W=1)=0 $ as this event cannot occur (this is true for any $w \notin \{2,3,4,5,6\} $).

A key observation to make is that the chips are replaced between drawings, so that each pull is independent of one another, i.e., the chance that I draw chip labeled with a 3 does not have any effect on which chip I draw next. This simplifies calculating the probabilities of each event since we can just multiply the probability of certain draws occurring.

I will calculate a few of these probabilities. For $P(W=2) $ we notice that the event that $W=2 $ can only occur in one way, that is, if $X=1 $ and $Y=1 $. We get

$$P(W=2) = P(X=1)P(Y=1) = \frac{30}{100}\cdot\frac{30}{100}=\frac{9}{100} $$

For $P(W=3) $, there are two ways of this occurring. We see that $ W=3$ if $X=1 $ and $Y=2 $ or if $X=2 $ and $Y=1$. Thus

$$P(W=3) = P(X=1)P(Y=2) + P(X=2)P(Y=1) = \frac{30}{100}\cdot\frac{20}{100} +\frac{20}{100}\cdot\frac{30}{100}=\frac{12}{100}. $$

Lastly, I will calculate $P(W=4) $ which can occur in three distinct ways,as Idiotic Shrike mentioned in the comments. That is, $X=1 $ and $Y=3 $, or $X=2 $ and $Y=2 $, or $X=3 $ and $Y=1 $. Therefore

$$ P(W=4) = P(X=1)P(Y=3) + P(X=2)P(Y=2)+P(X=3)P(Y=1)$$ $$ = \frac{30}{100}\cdot\frac{50}{100} +\frac{20}{100}\cdot\frac{20}{100}+\frac{50}{100}\cdot\frac{30}{100}=\frac{34}{100}$$

I will let you compute the rest.