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Question Number 138060 by mathocean1 last updated on 09/Apr/21

We throw 2 times a cubic dice . Its faces are numeroted 0, 0, −1, 1,1,1. We note a the result of the face throw and b the result of second one. 1)Calculate the probabilities to have a and b which verify : •a=−1 and b=0 •a=1 and b=0 •a^2 +b^2 =2 •a−b=1 2) Show that the probability to have a−b=1 knowing that a^2 +b^2 =2 is 0.75

$${We}\:{throw}\:\mathrm{2}\:{times}\:{a}\:{cubic}\:{dice}\:.\:{Its} \\ $$$${faces}\:{are}\:{numeroted}\:\mathrm{0},\:\mathrm{0},\:−\mathrm{1},\:\mathrm{1},\mathrm{1},\mathrm{1}. \\ $$$${We}\:{note}\:{a}\:{the}\:{result}\:{of}\:{the}\:{face}\:{throw} \\ $$$${and}\:{b}\:{the}\:{result}\:{of}\:{second}\:{one}.\: \\ $$$$\left.\mathrm{1}\right){Calculate}\:{the}\:{probabilities}\:{to}\:{have}\: \\ $$$${a}\:{and}\:{b}\:{which}\:{verify}\:: \\ $$$$\bullet{a}=−\mathrm{1}\:{and}\:{b}=\mathrm{0} \\ $$$$\bullet{a}=\mathrm{1}\:{and}\:{b}=\mathrm{0} \\ $$$$\bullet{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2} \\ $$$$\bullet{a}−{b}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{Show}\:{that}\:{the}\:{probability}\:{to}\:{have} \\ $$$${a}−{b}=\mathrm{1}\:{knowing}\:{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}\:{is}\:\mathrm{0}.\mathrm{75} \\ $$

Answered by floor(10²Eta[1]) last updated on 09/Apr/21

1) (1/6)×(2/6)=(1/(18)) (3/6)×(2/6)=(1/6) a^2 +b^2 =2⇒(a, b)=(1,1),(1,−1),(−1,1),(−1,−1) (4/6)×(4/6)=(4/9) a−b=1⇒(a, b)=(0, −1),(1,0) (2/6)×(1/6)+(3/6)×(2/6)=(2/9) 2)if a^2 +b^2 =2 so a and b can′t be 0 so a−b≠1. the probability is 0

$$\left.\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{2}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{18}} \\ $$$$\frac{\mathrm{3}}{\mathrm{6}}×\frac{\mathrm{2}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{2}\Rightarrow\left(\mathrm{a},\:\mathrm{b}\right)=\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{1},−\mathrm{1}\right),\left(−\mathrm{1},\mathrm{1}\right),\left(−\mathrm{1},−\mathrm{1}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{6}}×\frac{\mathrm{4}}{\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{a}−\mathrm{b}=\mathrm{1}\Rightarrow\left(\mathrm{a},\:\mathrm{b}\right)=\left(\mathrm{0},\:−\mathrm{1}\right),\left(\mathrm{1},\mathrm{0}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{3}}{\mathrm{6}}×\frac{\mathrm{2}}{\mathrm{6}}=\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\left.\mathrm{2}\right)\mathrm{if}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{2}\:\mathrm{so}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{0}\:\mathrm{so} \\ $$$$\mathrm{a}−\mathrm{b}\neq\mathrm{1}.\:\mathrm{the}\:\mathrm{probability}\:\mathrm{is}\:\mathrm{0} \\ $$

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