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Question Number 165669 by mkam last updated on 06/Feb/22

Commented by mkam last updated on 06/Feb/22

$$helpmemsrw$$

Answered by mr W last updated on 06/Feb/22

$$die1:1,2,3,4,5,6$$$$die2:1,2,3,4$$$$X=sumofbothdice$$$$(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!})(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!})$$$$=x^2+2x^3+3x^4+4x^5+4x^6+4x^7+3x^8+2x^9+x^{10}$$$$P(X=2)=P(X=10)=\frac{1}{24}$$$$P(X=3)=P(X=9)=\frac{2}{24}$$$$P(X=4)=P(X=8)=\frac{3}{24}$$$$P(X=5)=P(X=6)=P(X=7)=\frac{4}{24}$$$$E\left[X\right]=(2+10)\times \frac{1}{24}+(3+9)\times \frac{2}{24}+(4+8)\times \frac{3}{24}+(5+6+7)\times \frac{4}{24}$$$$=6$$

Commented by Tawa11 last updated on 06/Feb/22

$$\mathrm{Great}\mathrm{sir}$$

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