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I assume that the $Y_1$ and $Y_2$ are independent and can only take two values: $b$ or $c$. Further, I know that:

$$P(Y_1=b)=a \;\; \text{and} \;\; P(Y_1=c)=1-a$$

And similarly for $Y_2$:

$$P(Y_2=b)=a \;\; \text{and} \;\; P(Y_2=c)=1-a$$

I am being asked to produce the probability distribution of $Y_1 + Y_2$. I think that the answer will be:

$$ \begin{align} P(Y_1 + Y_2 = 2b) &= a^2 \\ P(Y_1 + Y_2 = b+c) &= a(1-a) \\ P(Y_1 + Y_2 = 2c) &= (1-a)^2 \end{align} $$

But it's just something that I believe would make sense. I have no good explanation for that result. My question is whether this result is correct and if so, could someone try to explain that to me or at least point me in the right direction?

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    $\begingroup$ If $Y_1$ takes a value of $b$ and $Y_2$ takes a value of $c$, what would it mean for $Y_1 + Y_2$ to take a value of $b+c$? $\endgroup$ – Dave Apr 19 at 17:07
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It’s correct except the second because it’s either $Y_1=b, Y_2=c$ or $Y_1=c, Y_2=b$. That makes $2a(1-a)$.

Your intution is correct, you’ll write the cases that lead to a specific outcome and accumulate the probabilities. Here, $$P(Y_1=b, Y_2=c)=P(Y_1=b)P(Y_2=c)=a(1-a)$$

A quick check: If you sum all three probabilies, it makes 1.

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