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-probab

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

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

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

$$\mathrm{Yes} \\ $$

Commented by MikeH last updated on 19/Mar/23

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

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

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

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