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

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

Commented by MikeH last updated on 19/Mar/23

$$\mathrm{please}\mathrm{my}\mathrm{main}\mathrm{worry}\mathrm{is},\mathrm{does}\mathrm{the}\mathrm{statement}$$$$‘‘\mathrm{A}\mathrm{man}\mathrm{has}\mathrm{equal}\mathrm{chances}\mathrm{of}\mathrm{travelling}\mathrm{by}$$$$\mathrm{air}\left(\mathrm{A}\right),\mathrm{bus}\left(\mathrm{B}\right)\mathrm{and}\mathrm{train}{\left(T\right)}^{″}\mathrm{means}$$$$\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)=\mathrm{P}\left(\mathrm{T}\right)=\frac{1}{3}?$$

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

$$letinvolvedinanaccident:E$$$$a)p\left(E\right)=p\left(A\right)p(R\mid A)+p\left(B\right)p(E\mid B)+p(E\mid T)=\frac{1}{3}(\frac{1}{3}+\frac{3}{5}+\frac{1}{10})=\frac{31}{90}$$$$b)p(A\mid E)=\frac{p(A\cap E)}{p\left(E\right)}=\frac{\frac{1}{9}}{\frac{31}{90}}=\frac{10}{31}$$$$c)p(B\cup E^c)=1-p(E-B)=1-p\left(E\right)+p(E\cap B)=\frac{23}{30}$$$$$$

Commented by MikeH last updated on 19/Mar/23

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

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