A computer producing factory has only two plants $${T_1}$$ and $${T_2}.$$ Plant $${T_1}$$ produces $$20$$% and plant $${T_2}$$ produces $$80$$% of the total computers produced. $$7$$% of computers produced in the factory turn out to be defective. It is known that $$P$$ (computer turns out to be defective given that it is produced in plant $${T_1}$$)
$$ = 10P$$ (computer turns out to be defective given that it is produced in plant $${T_2}$$),
where $$P(E)$$ denotes the probability of an event $$E$$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $${T_2}$$ is

The total number of distinct $$x \in \left[ {0,1} \right]$$ for which

$$\int\limits_0^x {{{{t^2}} \over {1 + {t^4}}}} dt = 2x - 1$$