Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

Consider three sets $E_1=\{1,2,3\},F_1=\{1,3,4\}$ and $G_1=\{2,3,4,5\}$. Two elements are chosen at random, without replacement, from the set $E_1$, and let $S_1$ denote the set of these chosen elements. Let $E_2=E_1-S_1$ and $F_2=F_1\cup S_1$. Now two elements are chosen at random, without replacement, from the set $F_2$ and let $S_2$ denote the set of these chosen elements.

Let $G_2=G_1\cup S_2$. Finally, two elements are chosen at random, without replacement, from the set $G_2$ and let $S_3$ denote the set of these chosen elements.
Let $E_3=E_2\cup S_3$. Given that $E_1=E_3$, let $p$ be the conditional probability of the event $S_1=\{1,2\}$. Then the value of $p$ is

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. The probability that a computer turns out to be defective which is produced in plant $T_1$ is ten times of the computers produced in the plant $T_2$. 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