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The number of computers of faculty members that are to be replaced in a certain department is 6.

The number of faculty members that have selected laptop machines is 2.

The number of faculty members that have selected desktop machines is 4.

The computers are numbered 1, 2,鈥�,6. Two computers are selected such that one outcome consists of computers 1 and 2, another consists of computers 1 and 3, and so on).

Let A be the event of faculty members that have selected laptop machines.

Let B be the event of faculty members that have selected desktop machines.

The sample space for the provided scenario can be represented as,

\(S = \left\{ \begin{aligned}\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\\\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right),\left( {4,5} \right),\left( {4,6} \right),\left( {5,6} \right)\end{aligned} \right\}\)

Where 1 and 2 represent the laptop machines and 3, 4, 5 and 6 represents the desktop machines.

Therefore, the total number of outcomes are 15.

a.

The outcomes that both selected setups are laptop machines is \(\left( {1,2} \right)\).

The number of possible outcome is 1.

The probability that both selected setups are for laptop computers is computed as,

\(\begin{aligned}P\left( {{\rm{two}}\;{\rm{laptop}}\;{\rm{computers}}} \right) &= \frac{{No.\;of\;possible\;outcomes}}{{Total\;number\;of\;outcomes}}\\ &= \frac{1}{{15}}\\ &= 0.067\end{aligned}\)

Therefore, the probability that both selected setups are for laptop computers is 0.067.

b.

The outcomes that both selected setups are desktop machines are,

\(\left\{ {\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right),\left( {4,5} \right),\left( {4,6} \right),\left( {5,6} \right)} \right\}\)

The number of possible outcomes is 6.

The probability that both selected setups are desktop machines is computed as,

\(\begin{aligned}P\left( {{\rm{two}}\;{\rm{desktop}}\;{\rm{computers}}} \right) &= \frac{{No.\;of\;possible\;outcomes}}{{Total\;number\;of\;outcomes}}\\ &= \frac{6}{{15}}\\ &= 0.4\end{aligned}\)

Therefore, the probability that both selected setups are desktop machines is 0.4.

c.

The outcomes that at least one selected setup is for a desktop computer are,

\(\left\{ \begin{aligned}\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\\\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right),\left( {4,5} \right),\left( {4,6} \right),\left( {5,6} \right)\end{aligned} \right\}\)

The number of possible outcomes is 14.

The probability that at least one selected setup is for a desktop computer is computed as,

\(\begin{aligned}P\left( {atleast\;one\;desktop\;computer} \right) &= \frac{{No.\;of\;possible\;outcomes}}{{Total\;number\;of\;outcomes}}\\ &= \frac{{14}}{{15}}\\ &= 0.933\end{aligned}\)

Therefore, the probability that at least one selected setup is for a desktop computer is 0.933.

d.

The outcomes that at least one computer of each type is chosen for setup are,

\(\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right)} \right\}\)

The number of possible outcomes is 8.

The probability that at least one computer of each type is chosen for setup is computed as,

\(\begin{aligned}P\left( {{\rm{at least one computer of each type}}} \right) &= \frac{{No.\;of\;possible\;outcomes}}{{Total\;number\;of\;outcomes}}\\ &= \frac{8}{{15}}\\ &= 0.533\end{aligned}\)

Therefore, the probability that at least one computer of each type is chosen for setup is 0.533.