EX 2.23 laser printer and inkjet printer

A university warehouse has received a shipment of 25 printers, of which 10 are laser printers and 15 are inkjet models. If 6 of these 25 are selected at random to be checked by a particular technician, what is the probability that exactly 3 of those selected are laser printers (so that the other 3 are inkjets)?

Let $D_3=\{$ exactly 3 of the 6 selected are inkjet printers $\}$. Assuming that any particular set of 6 printers is as likely to be chosen as is any other set of 6, we have equally likely outcomes, so $P\left(D_3\right)=N\left(D_3\right)/N$, where $N$ is the number of ways of choosing 6 printers from the 25 and $N\left(D_3\right)$ is the number of ways of choosing 3 laser printers and 3 inkjet models. Thus $N=\left(\begin{array}{c}25\\ 6\end{array}\right)$ . To obtain $N\left(D_3\right)$, think of first choosing 3 of the 15 inkjet models and then 3 of the laser printers. There are $\left(\begin{array}{c}15\\ 3\end{array}\right)$ ways of choosing the 3 inkjet models, and there are $\left(\begin{array}{c}10\\ 3\end{array}\right)$ ways of choosing the 3 laser printers; $N\left(D_3\right)$ is now the product of these two numbers (visualize a tree diagram-we are really using a product rule argument here), so $$ P\left( {D}{3}\right) = \frac{N\left( {D}{3}\right) }{N} = \frac{\left( \begin{matrix} {15} \ 3 \end{matrix}\right) \left( \begin{matrix} {10} \ 3 \end{matrix}\right) }{\left( \begin{matrix} {25} \ 6 \end{matrix}\right) } = \frac{\frac{{15}!}{3!{12}!} \cdot \frac{{10}!}{3!7!}}{\frac{{25}!}{6!{19}!}} = {.3083}

Let ${D}_{4} = \{$ exactly 4 of the 6 printers selected are inkjet models $\}$ and define ${D}_{5}$ and ${D}_{6}$ in an analogous manner. Then the probability that at least 3 inkjet printers are selected is

\begin{align*} &P\left( {{D}{3} \cup {D}{4} \cup {D}{5} \cup {D}{6}}\right) \ &= P\left( {D}{3}\right) + P\left( {D}{4}\right) + P\left( {D}{5}\right) + P\left( {D}{6}\right) \ &= \frac{\left( \begin{matrix} {15} \ 3 \end{matrix}\right) \left( \begin{matrix} {10} \ 3 \end{matrix}\right) }{\left( \begin{matrix} {25} \ 6 \end{matrix}\right) } + \frac{\left( \begin{matrix} {15} \ 4 \end{matrix}\right) \left( \begin{matrix} {10} \ 2 \end{matrix}\right) }{\left( \begin{matrix} {25} \ 6 \end{matrix}\right) } + \frac{\left( \begin{matrix} {15} \ 5 \end{matrix}\right) \left( \begin{matrix} {10} \ 1 \end{matrix}\right) }{\left( \begin{matrix} {25} \ 6 \end{matrix}\right) } + \frac{\left( \begin{matrix} {15} \ 6 \end{matrix}\right) \left( \begin{matrix} {10} \ 0 \end{matrix}\right) }{\left( \begin{matrix} {25} \ 6 \end{matrix}\right) } = {.8530} \end{align*}$$