Determining P(X = 1 | X + Y ≥ 0)

Question

Let $X$ denote a discrete random variable with PMF $P_X$ given by

$$ P_X(x) = \begin{cases} 0.2 &\text{ if x = −1,} \\ 0.3 &\text{ if x = 0,} \\ 0.5 &\text{ if x = 1,} \\ 0 &\text{ otherwise.} \end{cases} $$

Let $Y$ denote a discrete random variable that is independent of $X$ and has PMF $P_Y$ given by $$ P_Y(y) = \begin{cases} 0.3 &\text{ if y = −1,} \\ 0.4 &\text{ if y = 0,} \\ 0.3 &\text{ if y = 1,} \\ 0 &\text{ otherwise.} \end{cases} $$

How do I then calculate $P(X = 1 \mid X + Y \geq 0).$ if it was just $P(X = 1 \mid Y \geq 0)$ then I would be able to do it but I don't know how to do when it's like this: $P(X = 1 \mid X + Y \geq 0).$

So far i have only this: $$ P(X = 1 | X + Y ≥ 0) = \frac{P(X = 1 \cap X + Y ≥ 0)}{P(X + Y ≥ 0)} $$ But from here i can't calculate it like i normally would do and have tried many different things but can't find the right way to do it.

Answer 1

From the conditional probability we have the following: $$ P(X = 1 | X + Y ≥ 0) = \frac{P(X = 1 \cap X + Y ≥ 0)}{P(X + Y ≥ 0)} $$

We now calculate $P(X = 1 \cap X + Y ≥ 0)$ \begin{align} P(X = 1 \cap X + Y ≥ 0) &= P(X = 1, Y = -1) + P(X = 1, Y = 0) + P(X = 1, Y = 1) \\ &= (0.5 \cdot 0.3) + (0.5 \cdot 0.4) + (0.5 \cdot 0.3) \\ &= 0.5 \end{align}

Last but not least we now calculate $P(X + Y ≥ 0)$ \begin{align} P(X + Y ≥ 0) &= P(X = -1, Y = 1) + P(X = 0, Y = 0) + P(X = 1, Y = -1) + P(X = 1, Y = 0) + P(X = 0, Y = 1) + P(X = 1, Y = 1) \\ &= (0.2 \cdot 0.3) + (0.3 \cdot 0.4) + (0.5 \cdot 0.3) + (0.5 \cdot 0.4) + (0.3 \cdot 0.3) + (0.5 \cdot 0.3) \\ &= 0.77 \end{align}

Therefore we now get: \begin{align} P(X = 1 | X + Y ≥ 0) &= \frac{P(X = 1 \cap X + Y ≥ 0)}{P(X + Y ≥ 0)} \\ &= \frac{0.50}{0.77} \\ &\approx 0.649 \end{align}