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Question Number 176168 by nadovic last updated on 14/Sep/22

$$\mathrm{Two}\mathrm{fair}\mathrm{dice}\mathrm{are}\mathrm{rolled}\mathrm{once}.\mathrm{Let}X$$$$\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{representing}$$$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{numbers}\mathrm{that}\mathrm{show}$$$$\mathrm{up}\mathrm{on}\mathrm{the}\mathrm{two}\mathrm{dice}.\mathrm{Find}X.$$

Answered by aleks041103 last updated on 14/Sep/22

$$let{x}_{1}and{x}_{2}bethenumbersonthetwodice.$$$$thenX=x_1+x_2$$$$thentheprobabilityofgivenvalueXis:$$$$P\left(X\right)=\underset{x_1=1}{\sum ^6}p(x_1,X-x_1)=$$$$=\underset{x=1}{\sum ^6}p\left(x\right)p(X-x)=\frac{1}{36}\underset{x=1}{\sum ^6}\{\begin{array}{l}1,1⩽X-x⩽6\\ 0,else\end{array}$$$$1⩽X-x⩽6\Rightarrow 1-X⩽-x⩽6-X$$$$\Rightarrow X-6⩽x⩽X-1$$$$\Rightarrow P\left(X\right)=\frac{1}{36}\mid ([X-6,X-1]\cap [1,6]\cap \mathbb{N})\mid$$$$obv.P(X<2)=0andP(X>12)=0$$$$$$$$forX\in [2,6]:X-6\notin \mathbb{N}whileX-1\in \mathbb{N}and6>X-1>0$$$$\Rightarrow P(X\in [2,6])=\frac{1}{36}\mid [1,X-1]\cap [1,6]\cap \mathbb{N}\mid =$$$$=\frac{1}{36}\mid \{1,2,...,X-1\}\mid =\frac{X-1}{36}$$$$$$$$forX=7:P\left(7\right)=\frac{1}{36}\mid \{1,2,...,6\}\mid =\frac{1}{6}$$$$$$$$forX\in [8,12]:[X-6,X-1]\cap [1,6]=[X-6,6]$$$$\Rightarrow P(X\in [8,12])=\frac{1}{36}\mid \{X-6,X-5,...,6\}\mid =$$$$=\frac{1}{36}(13-X)$$$$\Rightarrow P\left(X\right)=\{\begin{array}{l}0,X\notin [2,12]\cap \mathbb{N}\\ \frac{X-1}{36},X\in [2,7]\cap \mathbb{N}\\ \frac{13-X}{36},X\in [8,12]\cap \mathbb{N}\end{array}$$

Commented by Tawa11 last updated on 15/Sep/22

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

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