Rolling a die - Conditional Probability

probability summation expected-value probability-distributions conditional-probability

$\mathbb{P}(X=1|Y=3)=\frac{1}{5},\mathbb{P}(X=2|Y=3)=\frac{4}{25},\mathbb{P}(X>3|Y=3)=1−(\frac{1}{5}+\frac{4}{25})=\frac{16}{25}.$

Then we multiply these by the expected results, i.e. $1,2$ and $9$, giving $\mathbb{E}(X|Y=3)=(1∗\frac{1}{5})+(2∗\frac{4}{25})+(9∗\frac{16}{25})=6.28$

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