Let-A-and-B-be-2-events-such-that-P-A-0-4-P-B-A-0-7-and-P-B-A-0-6-then-find-i-P-B-A-ii-P-A-B-iii-P-B-iv-P-A-B-

Question Number 145869 by Ar Brandon last updated on 09/Jul/21

$$\mathrm{Let}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{be}\:\mathrm{2}\:\mathrm{events}\:\mathrm{such}\:\mathrm{that}\:\mathrm{P}\left(\mathrm{A}\right)=\mathrm{0}.\mathrm{4},\:\mathrm{P}\left(\overset{−} {\mathrm{B}}/\mathrm{A}\right)=\mathrm{0}.\mathrm{7} \\ $$$$\:\mathrm{and}\:\mathrm{P}\left(\mathrm{B}/\overset{−} {\mathrm{A}}\right)=\mathrm{0}.\mathrm{6},\:\mathrm{then}\:\mathrm{find} \\ $$$$\left({i}\right)\mathrm{P}\left(\overset{−} {\mathrm{B}}/\overset{−} {\mathrm{A}}\right)\:\:\left({ii}\right)\mathrm{P}\left(\mathrm{A}\cap\mathrm{B}\right)\:\:\:\left({iii}\right)\:\mathrm{P}\left(\mathrm{B}\right)\:\:\left({iv}\right)\mathrm{P}\left(\mathrm{A}\cup\mathrm{B}\right) \\ $$

Answered by gsk2684 last updated on 09/Jul/21

using multiplication theorem P(A∩B^− )=0.4×0.7 ⇒P(A)−P(A∩B)=0.28 ⇒P(A∩B)=0.4−0.28=0.12 & P(B∩A^− )=0.6×0.6 ⇒P(B)−P(B∩A)=0.36 ⇒P(B)=0.36+0.12=0.48 proceed with these

$${using}\:{multiplication}\:{theorem}\: \\ $$$${P}\left({A}\cap\overset{−} {{B}}\right)=\mathrm{0}.\mathrm{4}×\mathrm{0}.\mathrm{7} \\ $$$$\Rightarrow{P}\left({A}\right)−{P}\left({A}\cap{B}\right)=\mathrm{0}.\mathrm{28} \\ $$$$\Rightarrow{P}\left({A}\cap{B}\right)=\mathrm{0}.\mathrm{4}−\mathrm{0}.\mathrm{28}=\mathrm{0}.\mathrm{12} \\ $$$$\&\:{P}\left({B}\cap\overset{−} {{A}}\right)=\mathrm{0}.\mathrm{6}×\mathrm{0}.\mathrm{6} \\ $$$$\Rightarrow{P}\left({B}\right)−{P}\left({B}\cap{A}\right)=\mathrm{0}.\mathrm{36} \\ $$$$\Rightarrow{P}\left({B}\right)=\mathrm{0}.\mathrm{36}+\mathrm{0}.\mathrm{12}=\mathrm{0}.\mathrm{48} \\ $$$${proceed}\:{with}\:{these} \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 09/Jul/21