Question

A box in a supply room contains \({\rm{15}}\) compact fluorescent lightbulbs, of which \({\rm{5}}\) are rated \({\rm{13}}\)-watt, \({\rm{6}}\)are rated \({\rm{18}}\)-watt, and \({\rm{4}}\) are rated \({\rm{23}}\)-watt. Suppose that three of these bulbs are randomly selected.

a. What is the probability that exactly two of the selected bulbs are rated \({\rm{23}}\)-watt?

b. What is the probability that all three of the bulbs have the same rating?

c. What is the probability that one bulb of each type is selected?

d. If bulbs are selected one by one until a \({\rm{23}}\)-watt bulb is obtained, what is the probability that it is necessary to examine at least 6 bulbs?

Short answer

a) The probability is \({\rm{0}}{\rm{.145}}{\rm{.}}\)

b) The probability is \({\rm{0}}{\rm{.075}}\)

c) The probability is \({\rm{0}}{\rm{.264}}\).

d) The probability is \({\rm{0}}{\rm{.154}}\).

Step-by-step solution

Definition

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Calculating the probability

We have total \({\rm{15}}\) compact fluorescent light bulbs, of which \({\rm{5}}\) are rated \({\rm{13}}\)-watt, 6 are rated \({\rm{18}}\)-watt, and \({\rm{4}}\) are rated \({\rm{23}}\) -watt and we select 3 randomly.

a)

In order to calculate the probability that exactly two of the selected bulbs are rated \({\rm{23}}\) -watt using

\({\rm{P(A) = }}\frac{{{\rm{ \# offavorableoutcomesinA}}}}{{{\rm{ \# ofoutcomesinthesamplespace}}}}\)

we need to find # of favorable outcomes in \({\rm{A}}\)and # outcomes in the sample space. We have four \({\rm{23}}\)-watt light bulbs, and the rest\({\rm{11(15 - 1 = 11}}\)) light bulbs are not \({\rm{23}}\) -watt. Using the combinations and the fundamental product rules:

Product Rule for k-Tuples

Assume that we can select first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways, and that for cache selected clement we can select the second element in \({{\rm{n}}_{\rm{2}}}\) ways, then we have the number of pairs to be \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\).

Similarly, for ordered collection of \({\rm{k}}\) elements, where the \({{\rm{k}}^{{\rm{th }}}}\)can be chosen in \({{\rm{n}}_{\rm{k}}}\) ways, then there are \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\) possible \({\rm{k}}\)-tuples.

Calculation

A combination is an unordered subset. For \({\rm{n}}\)individuals in a group, the number of combinations size \({\rm{k}}\)is denoted as \({{\rm{C}}_{{\rm{k,n}}}}\)or \(\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\)$ which we read as from \({\rm{n}}\)elements we choose\({\rm{k}}\). The proposition:

\(\begin{aligned}{{\rm{C}}_{{\rm{k,n}}}}\rm &= \left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\\\rm &= \frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}\\\rm &= \frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\end{aligned}\)

The number of ways to select two out of four light bulbs which are \({\rm{23}}\) -watt, denoted as\({{\rm{n}}_{\rm{1}}}\), is

\(\begin{aligned}{{\rm{n}}_{\rm{1}}}\rm &= {{\rm{C}}_{{\rm{2,4}}}}\\\rm &= \left( {\begin{aligned}{\rm{4}}\\{\rm{2}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{4!}}}}{{{\rm{2!(4 - 2)!}}}}\\\rm &= 6\end{aligned}\)

and the number of ways to select\({\rm{1}}\left( {{\rm{3 - 1 = 2}}} \right.\)) light bulbs out of \({\rm{11}}\) remaining, denoted as\({{\rm{n}}_{\rm{2}}}\), is

\(\begin{aligned}{{\rm{n}}_{\rm{2}}}\rm &= {{\rm{C}}_{{\rm{1,11}}}}\\\rm &= \left( {\begin{aligned}{{\rm{11}}}\\{\rm{1}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{11!}}}}{{{\rm{1!(11 - 1)!}}}}\\\rm &= 11\end{aligned}\)

Calculation

Therefore, the number of ways to select exactly two \({\rm{23}}\)-watt out of three randomly selected light bulbs is

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{ = 6*11 = 66}}\)

The size of the sample space, denoted as\({\rm{N}}\), and using combinations is

\(\begin{aligned}\rm N &= {{\rm{C}}_{{\rm{3,15}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{15}}}\\{\rm{3}}\end{aligned}} \right)\\\rm &= \frac{{{\rm{15!}}}}{{{\rm{3!(15 - 3)!}}}}{\rm{ = 455}}\end{aligned}\)

ways, where we actually chose \({\rm{3}}\) light bulbs out of \({\rm{15}}\) possible.

The probability is

\(\begin{aligned}\rm P(A) &= \frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\\\rm &= \frac{{{\rm{66}}}}{{{\rm{455}}}}\\\rm &= 0{\rm{.145}}{\rm{.}}\end{aligned}\)

Calculating the probability

We have total \({\rm{15}}\) compact fluorescent light bulbs, of which \({\rm{5}}\) are rated \({\rm{13}}\) -watt, \({\rm{6}}\)are rated \({\rm{18}}\) -watt, and \({\rm{4}}\) are rated \({\rm{23}}\) -watt and we select \({\rm{3}}\) randomly.

A combination is an unordered subset. For \({\rm{n}}\)individuals in a group, the number of combinations size \({\rm{k}}\) is denoted as \({{\rm{C}}_{{\rm{k,n}}}}\) or

\(\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\)

which we read as from \({\rm{n}}\)elements we choose\({\rm{k}}\). The proposition:

\(\begin{aligned}{{\rm{C}}_{{\rm{k,n}}}}\rm &= \left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\\\rm &= \frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}\\\rm &= \frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\end{aligned}\)

b)

In order to calculate the probability that all three of the bulbs have the same rating using

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\)

we need to find # of favorable outcomes in A and # of outcomes in the sample space.

\({\rm{513\;W}}\)bulbs, using combinations, is

\(\left( {\begin{aligned}{\rm{5}}\\{\rm{3}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{5!}}}}{{{\rm{3!(5 - 3)!}}}}{\rm{ = 10}}\)

Calculation

The number of ways to select \({\rm{3}}\) out of \({\rm{618\;W}}\)bulbs, using combinations, is

\(\left( {\begin{aligned}{\rm{6}}\\{\rm{3}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{6!}}}}{{{\rm{3!(6 - 3)!}}}}{\rm{ = 20}}{\rm{. }}\)

The number of ways to select \({\rm{3}}\) out of \({\rm{424\;W}}\) bulbs, using combinations, is

\(\left( {\begin{aligned}{\rm{4}}\\{\rm{3}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{4!}}}}{{{\rm{3!(4 - 3)!}}}}{\rm{ = 4}}\) The number of favorable outcomes is the sum \({\rm{10 + 20 + 4 = 34}}\) ways.

The size of the sample space, denoted as \({\rm{N}}\), and using combinations is

\(\begin{aligned}\rm N &= {{\rm{C}}_{{\rm{3,15}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{15}}}\\{\rm{3}}\end{aligned}} \right)\\\rm &= \frac{{{\rm{15!}}}}{{{\rm{3!(15 - 3)!}}}}{\rm{ = 455}}\end{aligned}\)

ways, where we actually chose \({\rm{3}}\) light bulbs out of \({\rm{15}}\) possible.

The probability is

\(\begin{aligned}\rm P(A) &= \frac{{{\rm{\# of favorable outcomes in A }}}}{{{\rm{\# of outcomes in the sample space }}}}\\\rm &= \frac{{{\rm{34}}}}{{{\rm{455}}}}\\\rm &= 0{\rm{.075}}{\rm{.}}\end{aligned}\)

Therefore, the probability is \({\rm{0}}{\rm{.075}}\).

Calculating the probability

c)

In order to calculate the probability that one bulb of each type is selected using

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\)

we need to find # of favorable outcomes in \({\rm{A}}\) and \({\rm{\# }}\) of outcomes in the sample space.

The number of favorable outcomes can be calculated using k-Tuples where \({{\rm{n}}_{\rm{1}}}\) is the number of ways to choose one \({\rm{13\;W}}\) bulb, \({{\rm{n}}_{\rm{2}}}\) is the number of ways to choose one \({\rm{18\;W}}\) bulb, and \({{\rm{n}}_{\rm{3}}}\) is the number of ways to choose one \({\rm{24\;W}}\) bulb. We can calculate that using combinations (it will obviously be \({\rm{5}}\) ways, \({\rm{6}}\) ways, \({\rm{4}}\) ways, respectively, since we choose only one of each type)

\(\begin{aligned}{{\rm{n}}_{\rm{1}}}{\rm{ = }}\left( {\begin{aligned}{\rm{5}}\\{\rm{1}}\end{aligned}} \right){\rm{ = 5}}\\{{\rm{n}}_{\rm{2}}}{\rm{ = }}\left( {\begin{aligned}{\rm{6}}\\{\rm{1}}\end{aligned}} \right){\rm{ = 6}}\\{{\rm{n}}_{\rm{3}}}{\rm{ = }}\left( {\begin{aligned}{\rm{4}}\\{\rm{1}}\end{aligned}} \right){\rm{ = 4}}\end{aligned}\)

therefore, the number of favorable outcomes is

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{*}}{{\rm{n}}_{\rm{3}}}{\rm{ = 5*6*4 = 120}}\)

ways.

Calculation

The size of the sample space, denoted as\({\rm{N}}\), and using combinations is

\(\begin{aligned}\rm N &= {{\rm{C}}_{{\rm{3,15}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{15}}}\\{\rm{3}}\end{aligned}} \right)\\\rm &= \frac{{{\rm{15!}}}}{{{\rm{3!(15 - 3)!}}}}\\\rm &= 455\end{aligned}\)

ways, where we actually chose \({\rm{3}}\) light bulbs out of \({\rm{15}}\) possible.

The probability is

\(\begin{aligned}\rm P(A) &= \frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\\\rm &= \frac{{{\rm{120}}}}{{{\rm{455}}}}\\\rm &= 0{\rm{.264}}{\rm{.}}\end{aligned}\)

Therefore, the probability is \({\rm{0}}{\rm{.264}}\).

Calculating the probability

d)

We have total \({\rm{15}}\) compact fluorescent light bulbs, of which \({\rm{5}}\) are rated \({\rm{13}}\) -watt, \({\rm{6}}\) are rated \({\rm{18}}\) -watt, and 4 are rated \({\rm{23}}\) -watt and we select \({\rm{3}}\) randomly. We will need the following definitions:

Event A - "it is necessary to examine at least \({\rm{6}}\) bulbs" can happen if and only if none of the first five bulbs is \({\rm{23W}}\) bulb. Now it is easier to calculate the probability. Using

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\)

we need to find # of favorable outcomes in A and # of outcomes in the sample space. The number of favorable outcomes can be calculated using combinations:

A combination is an unordered subset. For \({\rm{n}}\)individuals in a group, the number of combinations size \({\rm{k}}\) is denoted as \({{\rm{C}}_{{\rm{k,n}}}}\) or

\(\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\)which we read as from \({\rm{n}}\) elements we choose \({\rm{k}}\). The proposition:

\(\begin{aligned}{{\rm{C}}_{{\rm{k,n}}}}\rm &= \left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\\\rm &= \frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}\\\rm &= \frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\end{aligned}\)

Which means that we choose five out of remaining \({\rm{11(15 - 4 = 11)}}\) light bulbs, therefore the number of favorable outcomes is

\(\begin{aligned}{{\rm{C}}_{{\rm{5,11}}}}\rm &= \left( {\begin{aligned}{{\rm{11}}}\\{\rm{5}}\end{aligned}} \right)\\\rm &= \frac{{{\rm{11!}}}}{{{\rm{5!(11 - 5)!}}}}\\\rm &= 462\end{aligned}\)

Calculation

The size of the sample space, denoted as \({\rm{N}}\), and using combinations is

\(\begin{aligned}\rm N &= {{\rm{C}}_{{\rm{5,15}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{15}}}\\{\rm{5}}\end{aligned}} \right)\\\rm &= \frac{{{\rm{15!}}}}{{{\rm{5!(15 - 5)!}}}}\\\rm &= 3003\end{aligned}\)

ways, where we actually chose \({\rm{5}}\) light bulbs out of \({\rm{15}}\) possible.

The probability is

\(\begin{aligned}\rm P(A) &= \frac{{{\rm{\# of favorable outcomes in A }}}}{{{\rm{\# of outcomes in the sample space }}}}\\\rm &= \frac{{{\rm{462}}}}{{{\rm{3003}}}}\\\rm &= 0{\rm{.154}}{\rm{.}}\end{aligned}\)

Therefore, the probability is \({\rm{0}}{\rm{.154}}\).