Question

# A, B and C can solve $50%$, $60%$ and $70%$of the sums from a book. If one sum from that book is given to them to solve then what is the probability that the sum will be solved?

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Solution

## Finding the probability:Let E, F and G be the events defined as follows,E $=$A solves the sum,F $=$B solves the sum,G $=$C solves the sum.Clearly, E, F and G are independents events such that,$\text{P(E)=}\frac{50}{100}=\frac{1}{2}$, $\text{P(F)=}\frac{60}{100}=\frac{3}{5}$and $\text{P(G)=}\frac{70}{100}=\frac{7}{10}$,Therefore, the required probability, $=\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{\cup }\mathbf{F}\mathbf{\cup }\mathbf{G}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{P}\mathbf{\left(}\overline{)\mathbf{E}}\mathbf{\right)}\mathbf{×}\mathbf{P}\mathbf{\left(}\overline{)\mathbf{F}}\mathbf{\right)}\mathbf{×}\mathbf{P}\mathbf{\left(}\overline{)\mathbf{G}}\mathbf{\right)}\mathrm{}\left[\therefore \mathrm{E},\mathrm{F}\mathrm{and}\mathrm{G}\mathrm{are}\mathrm{independents}\mathrm{events}\right]\phantom{\rule{0ex}{0ex}}=1-\left(1-\frac{1}{2}\right)×\left(1-\frac{3}{5}\right)×\left(1-\frac{7}{10}\right)\phantom{\rule{0ex}{0ex}}=1-\frac{1}{2}×\frac{2}{5}×\frac{3}{10}\phantom{\rule{0ex}{0ex}}=1-\frac{6}{100}\phantom{\rule{0ex}{0ex}}=\frac{94}{100}\phantom{\rule{0ex}{0ex}}=0.94$Hence, the required probability is $0.94$.

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