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?

Answer 1

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,

$$\begin{gathered} =\mathbf{P}\mathbf{(}\mathbf{E}\mathbf{\cup }\mathbf{F}\mathbf{\cup }\mathbf{G}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{-}\mathbf{P}\mathbf{\left(}\overline{\mathbf{E}}\mathbf{\right)}\mathbf{\times }\mathbf{P}\mathbf{\left(}\overline{\mathbf{F}}\mathbf{\right)}\mathbf{\times }\mathbf{P}\mathbf{\left(}\overline{\mathbf{G}}\mathbf{\right)}\mathrm{}\left[\therefore \mathrm{E},\mathrm{F}\mathrm{and}\mathrm{G}\mathrm{are}\mathrm{independents}\mathrm{events}\right] \\ =1-\left(1-\frac{1}{2}\right)\times \left(1-\frac{3}{5}\right)\times \left(1-\frac{7}{10}\right) \\ =1-\frac{1}{2}\times \frac{2}{5}\times \frac{3}{10} \\ =1-\frac{6}{100} \\ =\frac{94}{100} \\ =0.94 \end{gathered}$$

Hence, the required probability is $0.94$.