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Three candidates solve a question. Odds in favour of the correct answer are $5: 2,4: 3$ and $3: 4$ respectively for the three candidates. What is the probability that at least two of them solve the question correctly?
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The correct answer is:
$209 / 343$
Odd in fav. for student $(\mathrm{A})=\frac{5}{2}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}\left(\mathrm{A}^{\prime}\right)}$
Odd in fav. for student $(\mathrm{B})=\frac{4}{3}=\frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}$
Odd in fav. for student $(\mathrm{C})=\frac{3}{4}=\frac{\mathrm{P}(\mathrm{C})}{\mathrm{P}\left(\mathrm{C}^{\prime}\right)}$
$\Rightarrow \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{2}{5} \mathrm{P}(\mathrm{A}), \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \mathrm{P}(\mathrm{B}), \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \mathrm{P}(\mathrm{C})$
Now $\mathrm{P}(\mathrm{A})+\mathrm{P}\left(\mathrm{A}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{A})+\frac{2}{5} \mathrm{P}(\mathrm{A})=1 \Rightarrow \mathrm{P}(\mathrm{A})=\frac{5}{7}$
Also $\mathrm{P}(\mathrm{B})+\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{B})+\frac{3}{4} \mathrm{P}(\mathrm{B})=1 \Rightarrow \mathrm{P}(\mathrm{B})=\frac{4}{7}$
And $\mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{C}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{C})+\frac{4}{3} \mathrm{P}(\mathrm{C})=1 \Rightarrow \mathrm{P}(\mathrm{C})=\frac{3}{7}$
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{2}{5} \times \frac{5}{7}=\frac{2}{7}, \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \times \frac{4}{7}=\frac{3}{7}$,
$\mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \times \frac{3}{7}=\frac{4}{7}$
Req. Prob. $=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right) \times \mathrm{P}(\mathrm{C})$
$+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})$
$=\frac{5}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{3}{7}+\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7}$
$=\frac{209}{343}$
Odd in fav. for student $(\mathrm{B})=\frac{4}{3}=\frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}$
Odd in fav. for student $(\mathrm{C})=\frac{3}{4}=\frac{\mathrm{P}(\mathrm{C})}{\mathrm{P}\left(\mathrm{C}^{\prime}\right)}$
$\Rightarrow \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{2}{5} \mathrm{P}(\mathrm{A}), \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \mathrm{P}(\mathrm{B}), \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \mathrm{P}(\mathrm{C})$
Now $\mathrm{P}(\mathrm{A})+\mathrm{P}\left(\mathrm{A}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{A})+\frac{2}{5} \mathrm{P}(\mathrm{A})=1 \Rightarrow \mathrm{P}(\mathrm{A})=\frac{5}{7}$
Also $\mathrm{P}(\mathrm{B})+\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{B})+\frac{3}{4} \mathrm{P}(\mathrm{B})=1 \Rightarrow \mathrm{P}(\mathrm{B})=\frac{4}{7}$
And $\mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{C}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{C})+\frac{4}{3} \mathrm{P}(\mathrm{C})=1 \Rightarrow \mathrm{P}(\mathrm{C})=\frac{3}{7}$
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{2}{5} \times \frac{5}{7}=\frac{2}{7}, \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \times \frac{4}{7}=\frac{3}{7}$,
$\mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \times \frac{3}{7}=\frac{4}{7}$
Req. Prob. $=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right) \times \mathrm{P}(\mathrm{C})$
$+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})$
$=\frac{5}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{3}{7}+\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7}$
$=\frac{209}{343}$
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