Fifteen telephones are received at a service center. Of these, $5$ are mobile, $6$ are cordless, and $4$ are wired. These $15$ phones are randomly numbered from $1$ to $15$ to establish the order in which they are serviced. Which of the following statement(s) is/are correct?
- The probability that among the first $3$ serviced, the first and third are mobile and the second is not, is $\dfrac{5 \times 10 \times 4}{15 \times 14 \times 13}.$
- The probability that the first four serviced are all the wired phones, is $\frac{1}{\left(\begin{array}{c}15 \\ 4\end{array}\right)}$.
- The probability that after servicing ten of these phones, only one of the three types remain to be serviced, is $\frac{\left(\begin{array}{c}6 \\ 5\end{array}\right)}{\left(\begin{array}{c}15 \\ 5\end{array}\right)}.$
- The probability that two phones of each type are among the first six serviced, is $\frac{\left(\begin{array}{l}5 \\ 2\end{array}\right)+\left(\begin{array}{l}6 \\ 2\end{array}\right)+\left(\begin{array}{l}4 \\ 2\end{array}\right)}{\left(\begin{array}{l}15 \\ 6\end{array}\right)}$.
