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Question Number 34445 by Rio Mike last updated on 06/May/18

A1) Events A and B are such that P(A) = (9/(30)), P(B)= (2/5) and P(A ∪ B) = (4/5) find a) P(A ∩ B) b) P(A∣B) c) P(B^� ) 2) A bag contains 3 black balls and 5 white balls. Two balls selected are random without replacement.Find the probability a) of selecting a black and white ball in order b) that the balls are of different colours c) the second ball is white

$$\left.\:{A}\mathrm{1}\right)\:\mathrm{Events}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:{P}\left({A}\right)\:=\:\frac{\mathrm{9}}{\mathrm{30}},\:{P}\left(\boldsymbol{{B}}\right)=\:\frac{\mathrm{2}}{\mathrm{5}}\:{and}\: \\ $$$${P}\left({A}\:\cup\:{B}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\left.{find}\:{a}\right)\:{P}\left({A}\:\cap\:{B}\right) \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:{b}\right)\:{P}\left({A}\mid{B}\right) \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:{c}\right)\:{P}\left(\bar {{B}}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{3}\:\mathrm{black}\:\mathrm{balls}\:\mathrm{and} \\ $$$$\mathrm{5}\:\mathrm{white}\:\mathrm{balls}.\:\mathrm{Two}\:\mathrm{balls}\:\mathrm{selected}\:\mathrm{are} \\ $$$$\mathrm{random}\:\mathrm{without}\:\mathrm{replacement}.\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{probability} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:{a}\right)\:{of}\:{selecting}\:{a}\:{black}\:{and}\:{white} \\ $$$${ball}\:{in}\:{order} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:{b}\right)\:{that}\:{the}\:{balls}\:{are}\:{of}\:{different}\:{colours} \\ $$$$\left.\:\:\:\:\:\:\:\:{c}\right)\:{the}\:{second}\:{ball}\:{is}\:{white}\: \\ $$

Commented by candre last updated on 06/May/18

2) a)P=(3/8)×(5/7)=((15)/(56))≅26,79% b)P=(3/8)×(5/7)+(5/8)×(3/7) =((2×15)/(56))=((15)/(28))≅53,57% c) P=(3/8)×(5/7)+(5/8)×(4/7) =((15+20)/(56)) =((35)/(56))=(5/8)=62,5%

$$\left.\mathrm{2}\right) \\ $$$$\left.{a}\right){P}=\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{5}}{\mathrm{7}}=\frac{\mathrm{15}}{\mathrm{56}}\cong\mathrm{26},\mathrm{79\%} \\ $$$$\left.{b}\right){P}=\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{5}}{\mathrm{7}}+\frac{\mathrm{5}}{\mathrm{8}}×\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$=\frac{\mathrm{2}×\mathrm{15}}{\mathrm{56}}=\frac{\mathrm{15}}{\mathrm{28}}\cong\mathrm{53},\mathrm{57\%} \\ $$$$\left.{c}\right) \\ $$$${P}=\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{5}}{\mathrm{7}}+\frac{\mathrm{5}}{\mathrm{8}}×\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$=\frac{\mathrm{15}+\mathrm{20}}{\mathrm{56}} \\ $$$$=\frac{\mathrm{35}}{\mathrm{56}}=\frac{\mathrm{5}}{\mathrm{8}}=\mathrm{62},\mathrm{5\%} \\ $$

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