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Question Number 76588 by Rio Michael last updated on 28/Dec/19
$$\mathrm{Given}\:\mathrm{three}\:\mathrm{events}\:\mathrm{A},\mathrm{B},\:\mathrm{and}\:\mathrm{C}\:\mathrm{sucb}\:\mathrm{that} \\ $$$$\mathrm{P}\left(\mathrm{A}\right)\:=\:\mathrm{P}\left(\mathrm{C}\right)\:,\:\mathrm{P}\left(\mathrm{A}\:\cap\:\mathrm{C}\right)\:=\:\frac{\mathrm{1}}{\mathrm{10}}\:,\:\mathrm{P}\left(\mathrm{A}\:\cup\:\mathrm{C}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{P}\left(\mathrm{C}\mid\mathrm{B}\right)\:=\:\frac{\mathrm{2}}{\mathrm{7}}\:\:,\:\mathrm{P}\left(\mathrm{B}\cup\mathrm{C}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{find}\: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{P}\left(\mathrm{A}\right) \\ $$$$\left.\mathrm{b}\right)\:\mathrm{P}\left(\mathrm{B}\right) \\ $$
Answered by john santu last updated on 28/Dec/19
$$\left.{a}\right)\:{consider}\:{P}\left({A}\cup{C}\right)=\:{P}\left({A}\right)+{P}\left({C}\right)−{P}\left({A}\cap{C}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2}{P}\left({A}\right)−\frac{\mathrm{1}}{\mathrm{10}}\rightarrow\:{P}\left({A}\right)\:=\frac{\mathrm{3}}{\mathrm{10}} \\ $$
Answered by john santu last updated on 28/Dec/19
$$\left.{b}\right){P}\left({B}\cup{C}\right)={P}\left({B}\right)+{P}\left({C}\right)−{P}\left({B}/{C}\right)×{P}\left({B}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{5}}={P}\left({B}\right)+\frac{\mathrm{3}}{\mathrm{10}}−\frac{\mathrm{2}}{\mathrm{7}}×{P}\left({B}\right) \\ $$$${P}\left({B}\right)=\left(\frac{\mathrm{8}}{\mathrm{10}}−\frac{\mathrm{3}}{\mathrm{10}}\right)×\frac{\mathrm{7}}{\mathrm{5}}=\frac{\mathrm{7}}{\mathrm{10}} \\ $$
Commented by Rio Michael last updated on 28/Dec/19
$$\mathrm{thanks} \\ $$