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Question Number 189641 by MikeH last updated on 19/Mar/23

A man has equal chances of travelling by air(A), bus(B) and train(T). the probability that when he travels by air, bus or train he will have an accident are (1/3), (3/5) and (1/(10)). find the probability that (a) he travelled and was involved in an accident (b) he was travelling by air given that he was involved in an accident. (c) he was travelling by bus or he arrived safely

$$\mathrm{A}\:\mathrm{man}\:\mathrm{has}\:\mathrm{equal}\:\mathrm{chances}\:\mathrm{of}\:\mathrm{travelling}\:\mathrm{by}\: \\ $$$$\mathrm{air}\left(\mathrm{A}\right),\:\mathrm{bus}\left(\mathrm{B}\right)\:\mathrm{and}\:\mathrm{train}\left(\mathrm{T}\right).\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{when}\:\mathrm{he}\:\mathrm{travels}\:\mathrm{by}\:\mathrm{air},\:\mathrm{bus}\:\mathrm{or}\:\mathrm{train}\:\mathrm{he} \\ $$$$\mathrm{will}\:\mathrm{have}\:\mathrm{an}\:\mathrm{accident}\:\mathrm{are}\:\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{10}}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{he}\:\mathrm{travelled}\:\mathrm{and}\:\mathrm{was}\:\mathrm{involved}\:\mathrm{in}\:\mathrm{an}\:\mathrm{accident} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{he}\:\mathrm{was}\:\mathrm{travelling}\:\mathrm{by}\:\mathrm{air}\:\mathrm{given}\:\mathrm{that}\:\mathrm{he}\:\mathrm{was} \\ $$$$\mathrm{involved}\:\mathrm{in}\:\mathrm{an}\:\mathrm{accident}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{he}\:\mathrm{was}\:\mathrm{travelling}\:\mathrm{by}\:\mathrm{bus}\:\mathrm{or}\:\mathrm{he}\:\mathrm{arrived}\:\mathrm{safely} \\ $$

Commented by MikeH last updated on 19/Mar/23

please my main worry is, does the statement “A man has equal chances of travelling by air(A), bus(B) and train(T)” means P(A) = P(B) = P(T) = (1/3)?

$$\mathrm{please}\:\mathrm{my}\:\mathrm{main}\:\mathrm{worry}\:\:\mathrm{is},\:\mathrm{does}\:\mathrm{the}\:\mathrm{statement}\: \\ $$$$``\mathrm{A}\:\mathrm{man}\:\mathrm{has}\:\mathrm{equal}\:\mathrm{chances}\:\mathrm{of}\:\mathrm{travelling}\:\mathrm{by}\: \\ $$$$\mathrm{air}\left(\mathrm{A}\right),\:\mathrm{bus}\left(\mathrm{B}\right)\:\mathrm{and}\:\mathrm{train}\left({T}\right)''\:\mathrm{means} \\ $$$$\mathrm{P}\left(\mathrm{A}\right)\:=\:\mathrm{P}\left(\mathrm{B}\right)\:=\:\mathrm{P}\left(\mathrm{T}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}? \\ $$

Commented by Frix last updated on 19/Mar/23

Yes

$$\mathrm{Yes} \\ $$

Commented by MikeH last updated on 19/Mar/23

alright thank you

$$\mathrm{alright}\:\mathrm{thank}\:\mathrm{you} \\ $$

Answered by mehdee42 last updated on 19/Mar/23

let involved in an accident:E a)p(E)=p(A)p(R∣A)+p(B)p(E∣B)+p(E∣T)=(1/3)((1/3)+(3/5)+(1/(10)))=((31)/(90)) b)p(A∣E)=((p(A∩E))/(p(E)))=((1/9)/((31)/(90)))=((10)/(31)) c)p(B∪E^c )=1−p(E−B)=1−p(E)+p(E∩B)=((23)/(30))

$${let}\:{involved}\:{in}\:{an}\:{accident}:{E} \\ $$$$\left.{a}\right){p}\left({E}\right)={p}\left({A}\right){p}\left({R}\mid{A}\right)+{p}\left({B}\right){p}\left({E}\mid{B}\right)+{p}\left({E}\mid{T}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{10}}\right)=\frac{\mathrm{31}}{\mathrm{90}} \\ $$$$\left.{b}\right){p}\left({A}\mid{E}\right)=\frac{{p}\left({A}\cap{E}\right)}{{p}\left({E}\right)}=\frac{\frac{\mathrm{1}}{\mathrm{9}}}{\frac{\mathrm{31}}{\mathrm{90}}}=\frac{\mathrm{10}}{\mathrm{31}} \\ $$$$\left.{c}\right){p}\left({B}\cup{E}^{{c}} \right)=\mathrm{1}−{p}\left({E}−{B}\right)=\mathrm{1}−{p}\left({E}\right)+{p}\left({E}\cap{B}\right)=\frac{\mathrm{23}}{\mathrm{30}} \\ $$$$ \\ $$

Commented by MikeH last updated on 19/Mar/23

thanks so much sir

$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$

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