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$\displaystyle \begin{aligned} P(Y=5)&=\sum \limits_{x=5}^6 P(Y=5 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{1024}+\frac{18}{4096} \biggr] \\&=\frac{22}{24576} \end{aligned}$

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$\displaystyle \begin{aligned}F_{Y_3}(y)&=P(Y_3 \le y) \\&= \sum_{j=3}^5 P(W =j) \\&=\binom{5}{3} \biggl(F(y)\biggr)^3 \biggl(1-F(y)\biggr)^2 + \binom{5}{4} \biggl(F(y)\biggr)^4 \biggl(1-F(y)\biggr)^1+ \binom{5}{5} \biggl(F(y)\biggr)^5 \biggl(1-F(y)\biggr)^0 \\&=\binom{5}{3} \biggl(\frac{y}{10}\biggr)^3 \biggl(1-\frac{y}{10}\biggr)^2 + \binom{5}{4} \biggl(\frac{y}{10}\biggr)^4 \biggl(1-\frac{y}{10}\biggr)^1+ \binom{5}{5} \biggl(\frac{y}{10}\biggr)^5 \biggl(1-\frac{y}{10}\biggr)^0 \\&=10 \biggl(\frac{y}{10}\biggr)^3 \biggl(1-\frac{y}{10}\biggr)^2 + 5 \biggl(\frac{y}{10}\biggr)^4 \biggl(1-\frac{y}{10}\biggr)^1+ \biggl(\frac{y}{10}\biggr)^5 \end{aligned}$

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