Let events be

E₁ = A is selected manager

E₂ = B is selected manager

E₃ = C is selected manager

E = radical change in the marketing strategy

P(E₁) = 

By Baye's theorem,

$\mathrm{P}({\mathrm{E}}_2/\mathrm{E}) = \frac{\mathrm{P}\left({\mathrm{E}}_2\right)\mathrm{P}(\mathrm{E}/{\mathrm{E}}_2)}{\mathrm{P}\left({\mathrm{E}}_1\right)\mathrm{P}(\mathrm{E}/{\mathrm{E}}_1) +\hspace{0.17em}\mathrm{P}\left({\mathrm{E}}_2\right)\mathrm{P}(\mathrm{E}/{\mathrm{E}}_2)+\mathrm{P}\left(\mathrm{E}3\right)\mathrm{P}(\mathrm{E}/{\mathrm{E}}_3)}$

As $\mathrm{P}(\mathrm{E}/{\mathrm{E}}_1) = 0.3, \mathrm{P}(\mathrm{E}/\mathrm{E}2) = 0.8, \mathrm{P}(\mathrm{E}/{\mathrm{E}}_3) = 0.5$

Applying Baye's theorem,

$\mathrm{P}({\mathrm{E}}_2/\mathrm{E}) = \frac{\left(1/7 \times 0.8\right)}{\left(4/7 \times 0.3 + 1/7 \times 0.8 +\hspace{0.17em}2 \times 0.5\right)}$

$\mathrm{P}({\mathrm{E}}_2/\mathrm{E}) = \frac{4}{15}$

$\mathrm{P}\left(\mathrm{Red}\right) = \frac{3}{6}, \mathrm{P}\left(\mathrm{Yellow}\right) = \frac{2}{6}, \mathrm{P}\left(\mathrm{Green}\right) = \frac{1}{6}$

The probability of drawing white ball,

 $\mathrm{P}\left(\frac{\mathrm{White}}{\mathrm{Red}}\right) = \frac{2}{5}, \mathrm{P}\left(\frac{\mathrm{White}}{\mathrm{Yellow}}\right) = \frac{4}{5}, \mathrm{P}\left(\frac{\mathrm{White}}{\mathrm{Green}}\right) = \frac{3}{7}$

$\mathrm{P}\left(\frac{\mathrm{Red}}{\mathrm{White}}\right) = \frac{\mathrm{P}\left(\mathrm{Red}\right) \times \mathrm{P}\left(\frac{\mathrm{White}}{\mathrm{Red}}\right)}{\mathrm{P}\left(\mathrm{Red}\right)\mathrm{P}\left({\displaystyle \frac{\mathrm{White}}{\mathrm{Red}}}\right) +\hspace{0.17em}\mathrm{P}\left(\mathrm{Yellow}\right)\mathrm{P}\left(\frac{\mathrm{White}}{\mathrm{Yellow}}\right) + \mathrm{P}\left(\mathrm{Green}\right)\mathrm{P}\left(\frac{\mathrm{White}}{\mathrm{Green}}\right)}$

             = $\frac{{\displaystyle \frac{3}{6} \times \frac{2}{5}}}{\left({\displaystyle \frac{3}{6} \times \frac{2}{5}}\right) +\left({\displaystyle \frac{2}{6} \times \frac{4}{5}}\right) +\hspace{0.17em}\left({\displaystyle \frac{1}{6}} \times {\displaystyle \frac{3}{7}}\right)}$

            = $\frac{\left({\displaystyle \frac{6}{30}}\right)}{{\displaystyle \frac{6}{30} + \frac{8}{30} +\hspace{0.17em}\frac{3}{42}}}$

            = $\frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{1}{5} + \frac{4}{15} +\hspace{0.17em}\frac{1}{14}}}$

           = $\frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{42 +\hspace{0.17em}56 +\hspace{0.17em}15}{210}}}$

           = (1/5) x (210/113)

           = 42/113 = 0.37

Given, n = 5, p = P(1, 3 , 5) = $\frac{1}{2}$, q = 1 - p = 1 - $\frac{1}{2} = \frac{{\displaystyle 1}}{{\displaystyle 2}}$

P(Maximum three success) = 1 -[P(4) + P(5)]

                                       = $1 - \left[5_{{\mathrm{C}}_4}{\left(\frac{1}{2}\right)}^4\left(\frac{1}{2}\right) + 5_{{\mathrm{C}}_5}{\left(\frac{1}{2}\right)}^5\right]$

                                       = $1 -\frac{1}{32} \left[5_{{\mathrm{C}}_4} +\hspace{0.17em}{5}_{{\mathrm{C}}_5}\right]$

                                       = $1 - \frac{1}{32}\left(5 + 1\right)$

                                       = $1 - \frac{3}{16}$

                                      = $\frac{13}{16} = 0.81$

Given, P(A) = $\frac{1}{2}$, P(B) = $\frac{1}{\mathrm{3,}}$ $\mathrm{P}(\mathrm{A}\cap \mathrm{B}) = \frac{1}{4}$

$\therefore$ P(A/B) = $\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)} = \frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{3}}} = \frac{3}{4}$

$\therefore$ P(B) = $\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)} = \frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{2}}} = \frac{1}{2}$

Let A win the race be E₁

B win the race be E₂

P(E₁) = $\frac{1}{3}$, P(E₂) = $\frac{1}{6}$

$\mathrm{P}(\mathrm{E}{\text{'}}_{1 }\cap \mathrm{E}{\text{'}}_2) = \mathrm{P}\left(\mathrm{E}{\text{'}}_{1 }\right).\mathrm{P}\left(\mathrm{E}{\text{'}}_2\right)$

                 = [1 - P(E₁)].[1 - P(E₂)]

                 = [1 - $\frac{1}{3}$].[1 - $\frac{1}{6}$]

                 = $\frac{2}{3} \times \frac{5}{6}$

                 = $\frac{5}{9}$

Thus, the probability of neither of them winning the race is 5/9.