Question Number 165669 by mkam last updated on 06/Feb/22
Commented by mkam last updated on 06/Feb/22
$${help}\:{me}\:{msr}\:{w} \\ $$
Answered by mr W last updated on 06/Feb/22
![die 1: 1,2,3,4,5,6 die 2: 1,2,3,4 X=sum of both dice ((x/(1!))+(x^2 /(2!))+(x^3 /(3!))+(x^4 /(4!)))((x/(1!))+(x^2 /(2!))+(x^3 /(3!))+(x^4 /(4!))+(x^5 /(5!))+(x^6 /(6!))) =x^2 +2x^3 +3x^4 +4x^5 +4x^6 +4x^7 +3x^8 +2x^9 +x^(10) P(X=2)=P(X=10)=(1/(24)) P(X=3)=P(X=9)=(2/(24)) P(X=4)=P(X=8)=(3/(24)) P(X=5)=P(X=6)=P(X=7)=(4/(24)) E[X]=(2+10)×(1/(24))+(3+9)×(2/(24))+(4+8)×(3/(24))+(5+6+7)×(4/(24)) =6](https://www.tinkutara.com/question/Q165679.png)
$${die}\:\mathrm{1}:\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6} \\ $$$${die}\:\mathrm{2}:\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$${X}={sum}\:{of}\:{both}\:{dice} \\ $$$$\left(\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}\right)\left(\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}\right) \\ $$$$={x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{6}} +\mathrm{4}{x}^{\mathrm{7}} +\mathrm{3}{x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{9}} +{x}^{\mathrm{10}} \\ $$$${P}\left({X}=\mathrm{2}\right)={P}\left({X}=\mathrm{10}\right)=\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${P}\left({X}=\mathrm{3}\right)={P}\left({X}=\mathrm{9}\right)=\frac{\mathrm{2}}{\mathrm{24}} \\ $$$${P}\left({X}=\mathrm{4}\right)={P}\left({X}=\mathrm{8}\right)=\frac{\mathrm{3}}{\mathrm{24}} \\ $$$${P}\left({X}=\mathrm{5}\right)={P}\left({X}=\mathrm{6}\right)={P}\left({X}=\mathrm{7}\right)=\frac{\mathrm{4}}{\mathrm{24}} \\ $$$${E}\left[{X}\right]=\left(\mathrm{2}+\mathrm{10}\right)×\frac{\mathrm{1}}{\mathrm{24}}+\left(\mathrm{3}+\mathrm{9}\right)×\frac{\mathrm{2}}{\mathrm{24}}+\left(\mathrm{4}+\mathrm{8}\right)×\frac{\mathrm{3}}{\mathrm{24}}+\left(\mathrm{5}+\mathrm{6}+\mathrm{7}\right)×\frac{\mathrm{4}}{\mathrm{24}} \\ $$$$=\mathrm{6} \\ $$
Commented by Tawa11 last updated on 06/Feb/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$