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Chapter 2: Q23E (page 66)

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered1, 2,鈥�, 6, then one outcome consists of computers 1 and2, another consists of computers 1 and 3, and so on).

a. What is the probability that both selected setups are for laptop computers?

b. What is the probability that both selected setups are desktop machines?

c. What is the probability that at least one selected setup is for a desktop computer?

d. What is the probability that at least one computer of each type is chosen for setup?

Short Answer

Expert verified

a. The probability that both selected setups are for laptop computers is 0.067.

b. The probability that both selected setups are desktop machines is 0.4.

c. The probability that at least one selected setup is for a desktop computer is 0.933.

d. The probability that at least one computer of each type is chosen for setup is 0.533

Step by step solution

01

Given information

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).

02

Construct the sample space

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.

03

Compute the probability

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.

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Most popular questions from this chapter

Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a \({\rm{CD}}\) player. Let \({{\rm{A}}_{\rm{1}}}\) be the event that the receiver functions properly throughout the warranty period, \({{\rm{A}}_{\rm{2}}}\) be the event that the speakers function properly throughout the warranty period, and \({{\rm{A}}_{\rm{3}}}\) be the event that the \({\rm{CD}}\) player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with \({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = }}{\rm{.95}}\), \({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}{\rm{.98}}\), and \({\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}{\rm{.80}}\).

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f. What is the probability that all three components function properly throughout the warranty period but that at least one fails within a month after the warranty expires?

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