Continuous Random Variable Conditional Proability

Question

I need to answer two questions:

1. Find $P(Y|X)$;
2. $P(0<Y<1/2 | X=0.15)$.

For #1 I know I would have to use the double integral and find Pxy and I understand how to do #1.

However, I'm completely stuck on #2 and don't understand how to use the value of $X= 0.15$ because if this is plugged into the conditional probability formula would it not make the denominator $0$?

Any help would be greatly appreciated.

Answer 1

$f(x|y) = \frac{f(x,y)}{f(y)}$, which happens to be $1$, i.e. $f(x|y)=f(x)$, so $X$ and $Y$ are independent. Therefore. $P(Y<a|X)=P(Y<a)$.

Answer 2

However, I'm completely stuck on #2 and don't understand how to use the value of $X=0.15$ because if this is plugged into the conditional probability formula would it not make the denominator $0$?

The probability density function is not zero at that value. The formula for this probability mass is:

$$\mathsf P(0\leq Y\leq1/2\mid X{=}0.15)=\dfrac{\int_0^{1/2}f_{\small X,Y}(0.15,y)~\mathrm d y}{\int_0^{1}f_{\small X,Y}(0.15,y)~\mathrm d y}$$

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For #1 I know I would have to use the double integral and find Pxy and I understand how to do #1.

No, you just need a single integral. Also, as you are looking for a conditional pdf, you should not use the symbol $P(Y\mid X)$, butt: $$f_{\small Y\mid X}(y\mid x)=\dfrac{f_{\small X,Y}(x,y)}{\int_0^1 f_{\small X,Y}(x,y)~\mathsf dy}$$