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 be the numbers on the two dice. then X=x_1 +x_2 then the probability of given value X is: P(X)=Σ_(x_1 =1) ^6 p(x_1 ,X−x_1 )= =Σ_(x=1) ^6 p(x)p(X−x)=(1/(36))Σ_(x=1) ^6 { ((1, 1≤X−x≤6)),((0, else)) :} 1≤X−x≤6⇒1−X≤−x≤6−X ⇒X−6≤x≤X−1 ⇒P(X)=(1/(36))∣([X−6,X−1]∩[1,6]∩N)∣ obv. P(X<2)=0 and P(X>12)=0 for X∈[2,6]: X−6∉N while X−1∈N and 6>X−1>0 ⇒P(X∈[2,6])=(1/(36))∣[1,X−1]∩[1,6]∩N∣= =(1/(36))∣{1,2,...,X−1}∣=((X−1)/(36)) for X=7:P(7)=(1/(36))∣{1,2,...,6}∣=(1/6) for X∈[8,12]: [X−6,X−1]∩[1,6]=[X−6,6] ⇒P(X∈[8,12])=(1/(36))∣{X−6,X−5,...,6}∣= =(1/(36))(13−X) ⇒P(X)= { ((0, X∉[2,12]∩N)),((((X−1)/(36)), X∈[2,7]∩N)),((((13−X)/(36)), X∈[8,12]∩N)) :}](https://www.tinkutara.com/question/Q176177.png)
$${let}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{be}\:{the}\:{numbers}\:{on}\:{the}\:{two}\:{dice}. \\ $$$${then}\:{X}={x}_{\mathrm{1}} +{x}_{\mathrm{2}} \\ $$$${then}\:{the}\:{probability}\:{of}\:{given}\:{value}\:{X}\:{is}: \\ $$$${P}\left({X}\right)=\underset{{x}_{\mathrm{1}} =\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{p}\left({x}_{\mathrm{1}} ,{X}−{x}_{\mathrm{1}} \right)= \\ $$$$=\underset{{x}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{p}\left({x}\right){p}\left({X}−{x}\right)=\frac{\mathrm{1}}{\mathrm{36}}\underset{{x}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\begin{cases}{\mathrm{1},\:\mathrm{1}\leqslant{X}−{x}\leqslant\mathrm{6}}\\{\mathrm{0},\:{else}}\end{cases} \\ $$$$\:\mathrm{1}\leqslant{X}−{x}\leqslant\mathrm{6}\Rightarrow\mathrm{1}−{X}\leqslant−{x}\leqslant\mathrm{6}−{X} \\ $$$$\Rightarrow{X}−\mathrm{6}\leqslant{x}\leqslant{X}−\mathrm{1} \\ $$$$\Rightarrow{P}\left({X}\right)=\frac{\mathrm{1}}{\mathrm{36}}\mid\left(\left[{X}−\mathrm{6},{X}−\mathrm{1}\right]\cap\left[\mathrm{1},\mathrm{6}\right]\cap\mathbb{N}\right)\mid \\ $$$${obv}.\:{P}\left({X}<\mathrm{2}\right)=\mathrm{0}\:{and}\:{P}\left({X}>\mathrm{12}\right)=\mathrm{0} \\ $$$$ \\ $$$${for}\:{X}\in\left[\mathrm{2},\mathrm{6}\right]:\:{X}−\mathrm{6}\notin\mathbb{N}\:{while}\:{X}−\mathrm{1}\in\mathbb{N}\:{and}\:\mathrm{6}>{X}−\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow{P}\left({X}\in\left[\mathrm{2},\mathrm{6}\right]\right)=\frac{\mathrm{1}}{\mathrm{36}}\mid\left[\mathrm{1},{X}−\mathrm{1}\right]\cap\left[\mathrm{1},\mathrm{6}\right]\cap\mathbb{N}\mid= \\ $$$$=\frac{\mathrm{1}}{\mathrm{36}}\mid\left\{\mathrm{1},\mathrm{2},…,{X}−\mathrm{1}\right\}\mid=\frac{{X}−\mathrm{1}}{\mathrm{36}} \\ $$$$ \\ $$$${for}\:{X}=\mathrm{7}:{P}\left(\mathrm{7}\right)=\frac{\mathrm{1}}{\mathrm{36}}\mid\left\{\mathrm{1},\mathrm{2},…,\mathrm{6}\right\}\mid=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$${for}\:{X}\in\left[\mathrm{8},\mathrm{12}\right]:\:\left[{X}−\mathrm{6},{X}−\mathrm{1}\right]\cap\left[\mathrm{1},\mathrm{6}\right]=\left[{X}−\mathrm{6},\mathrm{6}\right] \\ $$$$\Rightarrow{P}\left({X}\in\left[\mathrm{8},\mathrm{12}\right]\right)=\frac{\mathrm{1}}{\mathrm{36}}\mid\left\{{X}−\mathrm{6},{X}−\mathrm{5},…,\mathrm{6}\right\}\mid= \\ $$$$=\frac{\mathrm{1}}{\mathrm{36}}\left(\mathrm{13}−{X}\right) \\ $$$$\Rightarrow{P}\left({X}\right)=\begin{cases}{\mathrm{0},\:{X}\notin\left[\mathrm{2},\mathrm{12}\right]\cap\mathbb{N}}\\{\frac{{X}−\mathrm{1}}{\mathrm{36}},\:{X}\in\left[\mathrm{2},\mathrm{7}\right]\cap\mathbb{N}}\\{\frac{\mathrm{13}−{X}}{\mathrm{36}},\:{X}\in\left[\mathrm{8},\mathrm{12}\right]\cap\mathbb{N}}\end{cases} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$