Probability and Statistics Questions

Question Number 110826 by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{of} \\$$$$\mathrm{the}\:\mathrm{events}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{occurs}\:\mathrm{is}\:\mathrm{0}.\mathrm{7}\:\mathrm{and} \\$$$$\mathrm{they}\:\mathrm{occur}\:\mathrm{simultaneously}\:\mathrm{with} \\$$$$\mathrm{probability}\:\mathrm{0}.\mathrm{2}.\:\mathrm{Then}\:\mathrm{P}\left(\overset{−} {\mathrm{A}}\right)+\mathrm{P}\left(\bar {\mathrm{B}}\right)\:= \\$$

Commented by kaivan.ahmadi last updated on 30/Aug/20

$${p}\left({A}\cup{B}\right)=\mathrm{0}.\mathrm{7} \\$$$${p}\left({A}\cap{B}\right)=\mathrm{0}.\mathrm{2} \\$$$${p}\left({A}\cup{B}\right)={p}\left({A}\right)+{p}\left({B}\right)−{p}\left({A}\cap{B}\right)\Rightarrow \\$$$$\mathrm{0}.\mathrm{7}={p}\left({A}\right)+{p}\left({B}\right)−\mathrm{0}.\mathrm{2}\Rightarrow \\$$$${p}\left({A}\right)+{p}\left({B}\right)=\mathrm{0}.\mathrm{9} \\$$$$\left(\mathrm{1}−{p}\left({A}\right)\right)+\left(\mathrm{1}−{p}\left({B}\right)\right)=−\mathrm{0}.\mathrm{9}+\mathrm{2} \\$$$$\Rightarrow{p}\left(\overset{−} {{A}}\right)+{p}\left(\overset{−} {{B}}\right)=\mathrm{1}.\mathrm{1} \\$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{the}\:\mathrm{question}\:\mathrm{asks}\:\mathrm{us}\:\mathrm{to}\:\mathrm{find} \\$$$$\mathrm{P}\left(\overset{−} {\mathrm{A}}\right)+\mathrm{P}\left(\overset{−} {\mathrm{B}}\right). \\$$

Commented by Aina Samuel Temidayo last updated on 31/Aug/20

$$\mathrm{Yes}. \\$$

Commented by Aina Samuel Temidayo last updated on 31/Aug/20

$$\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{sum}.\:\mathrm{So}\:\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{accepted}. \\$$

Commented by kaivan.ahmadi last updated on 31/Aug/20

$$\mathrm{0}\leqslant{p}\left(\overset{−} {{A}}\right)\leqslant\mathrm{1} \\$$$$\mathrm{0}\leqslant{p}\left(\overset{−} {{B}}\right)\leqslant\mathrm{1} \\$$$$\mathrm{0}\leqslant{p}\left(\overset{−} {{A}}\right)+{p}\left(\overset{−} {{B}}\right)\leqslant\mathrm{2} \\$$

Commented by Her_Majesty last updated on 31/Aug/20

$${P}\left(\overset{−} {{A}}\right)+{P}\left(\overset{−} {{B}}\right)\neq{P}\left(\overset{−} {{A}}\wedge\overset{−} {{B}}\right) \\$$$${the}\:{sum}\:{of}\:\mathrm{2}\:{independent}\:{probabilities}\:{is} \\$$$${a}\:{senseless}\:{value}. \\$$$${i}.{e}.\:{the}\:{probability}\:{of}\:{a}\:{thrown}\:{dice}\:{showing} \\$$$${not}\:\mathrm{6}\:{is}\:\frac{\mathrm{5}}{\mathrm{6}},\:{the}\:{one}\:{of}\:{a}\:{thrown}\:{coin}\:{showing} \\$$$${head}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{the}\:{sum}\:{is}\:\frac{\mathrm{4}}{\mathrm{3}}\approx\mathrm{133\%} \\$$