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

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