Calculation of conditional probability for an expression

Question

I have this question on conditional probability

A biased four-sided die is rolled, and the down face is a random variable $X$, described by the following probability mass function :

$$p_{\small X}(x)=\begin{cases}x/10&:& x\in\{1,2,3,4\}\\0&:&\textsf{otherwise}\end{cases}$$

Given the random variable $X$ a biased coin is flipped, and the random variable $Y$ is one or zero according to the result. The conditional mpf is

$$p_{\small Y\mid X}(y\mid x) = \left(\tfrac{x+1}{2x}\right)^y\left(1-\tfrac {x+1}{2x}\right)^{1-y}\quad[y\in\{0,1\} $$

The conditional probability formula is

$$p_{\small X\mid Y}({x}\mid{y})=\frac{p_{\small Y\mid X}({y}\mid{x})\cdot p_{\small X}(x)}{p_{\small Y}(y)}$$

Now, $p_{\small Y\mid X}({y}\mid{x})=p_{\small Y\mid X}({0}\mid{x})+p_{\small Y\mid X}({1}\mid{x})$ where $x\in\{1,2,3,4\}$ and $p_{\small X}(x)=\frac{x}{10}$ at $x\in\{1,2,3,4\}$

My question is how to calculate P(Y) as Y takes {0,1}

Answer 1

Now, $p_{\small Y\mid X}({y}\mid{x})=p_{\small Y\mid X}({0}\mid{x})+p_{\small Y\mid X}({1}\mid{x})$ where $x\in\{1,2,3,4\}$ and $p_{\small X}(x)=\frac{x}{10}$ at $x\in\{1,2,3,4\}$

No.   $p_{\small Y\mid X\!}(y\mid x) = \left(\tfrac{x+1}{2x}\right)^y\left(1-\tfrac {x+1}{2x}\right)^{1-y}\quad[\textsf{where }y\in\{0,1\}, x\in\{1,2,3,4\}]$

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Note: this means that $p_{\small Y\mid X}(0\mid x)=(\tfrac{x-1}{2x})$ and $p_{\small Y\mid X}(1\mid x)=(\tfrac{x+1}{2x})$ , for all $x\in\{1,2,3,4\}$

These do not sum to give $p_{\small Y\mid X\!}(y\mid x)$.   They sum to give $1$, as a well behaved probability mass function should.

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My question is how to calculate P(Y) as Y takes {0,1}

The law of total probability says: when $y\in\{0,1\}$ then:

$$\begin{align}p_{\small Y\!}(y) &= \sum_{x\in\{1,2,3,4\}} p_{\small Y\mid X\!}(y\mid x)\,p_{\small X\!}(x)\\[2ex]&= {p_{\small Y\mid X\!}(y\mid 1)\,p_{\small X\!}(1)}+{p_{\small Y\mid X\!}(y\mid 2)\,p_{\small X\!}(2)}+{p_{\small Y\mid X\!}(y\mid 3)\,p_{\small X\!}(3)}+{p_{\small Y\mid X\!}(y\mid 4)\,p_{\small X\!}(4)}\\[2ex]&=\sum_{x\in\{1,2,3,4\}}\left(\tfrac{x+1}{2x}\right)^y\left(\tfrac {x-1}{2x}\right)^{1-y}\tfrac x{10}\\&=\tfrac 1{20}\sum_{x\in\{1,2,3,4\}}(x+1)^y(x-1)^{1-y}\qquad[\textsf{where }y\in\{0,1\}]\\[2ex]&~~\vdots \end{align}$$

Answer 2

My question is how to calculate P(Y) as Y takes {0,1}

$$p(Y=y)=\sum_{x=1}^4p(x, y)$$

Thus $Y$ takes the values $\{0;1\}$ with probabilities $\{6/20;14/20\}$ respectively