Probability Test - 44

$${\textbf{Step - 1: Writing out different events and probabilities}}{\text{.}}$$

                  $${\text{Let A be an event that the car met with an accident}}$$

                  $$\text{Let}$$ $$\text{B}_1$$ $$\text{be an event where the driver was alcoholic}\text{.}$$

                  $${\text{And }}{{\text{B}}_2}{\text{ be an event where the drives was not alcoholic}}{\text{.}}$$

                  $${\text{P(}}{{\text{B}}_1}{\text{) = }}\dfrac{1}{5}{\text{    P(}}{{\text{B}}_2}{\text{) = 1 - }}\dfrac{1}{5} = \dfrac{4}{5}.$$

$${\textbf{Step - 2: Finding probability that the driver was alcoholic and the accident occured}}{\text{.}}$$

                  $${\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{B}}_1}}}} \right) = 0.001$$

                  $${\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{B}}_2}}}} \right) = 0.0001$$

                  $${\text{Using Bayes' theorem we can write, }}$$

                  $${\text{P}}\left( {\dfrac{{{{\text{B}}_1}}}{{\text{A}}}} \right) = \dfrac{{{\text{P(}}{{\text{B}}_1}{\text{)}} \times {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{B}}_1}}}} \right)}}{{{\text{P(}}{{\text{B}}_1}{\text{)}} \times {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{B}}_1}}}} \right) + {\text{P(}}{{\text{B}}_2}{\text{)}} \times {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{B}}_2}}}} \right)}}$$

                  $${\text{ = }}\dfrac{{0.2 \times 0.001}}{{0.2 \times .001 + 0.8 \times 0.0001}}$$

                  $$ = \dfrac{{20}}{{28}} = \dfrac{5}{7}.$$

$${\textbf{Thus, we get the probability to be }}\mathbf{\dfrac{5}{7}.}$$