Museum management. Refer to the Museum Management and Curatorship (June 2010) study of the criteria used to evaluate museum performance, Exercise 2.14 (p. 74). Recall that the managers of 30 leading museums of contemporary art were asked to provide the performance measure used most often. A summary of the results is reproduced in the table. Performance Measure Number of Museums Total visitors 8 Paying visitors 5 Big shows 6 Funds raised 7 Members 4

Performance Measure	Number of Museums
Total visitors	8
Paying visitors	5
Big shows	6
Funds raised	7
Members	4

a. If one of the 30 museums is selected at random, what is the probability that the museum uses total visitors or funds raised most often as a performance measure?

b. Consider two museums of contemporary art randomly selected from all such museums. Of interest is whether or not the museums use total visitors or funds raised most often as a performance measure. Use a tree diagram to aid in listing the sample points for this problem.

c. Assign reasonable probabilities to the sample points of part b.

d. Refer to parts b and c. Find the probability that both museums use total visitors or funds raised most often as a performance measure.

Let 'A' be the event in which a randomly selected museum utilizes total visitors as a performance measure the most often, and 'B' be the event in which funds raised are used as a performance measure. The probability that the museum utilizes total visitors or funds raised as a performance measure is:

$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}$

As the Number of museums using total visitors as a performance measure = 8, So:

$\begin{array}{c}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{8}}{\mathbf{30}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}}{\mathbf{15}}\end{array}$

As some museums use funds raised as performance measure = 7, So:

$\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\frac{\mathbf{7}}{\mathbf{30}}$

Therefore,

$\begin{array}{c}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{4}}{\mathbf{15}}\mathbf{+}\frac{\mathbf{7}}{\mathbf{30}}\mathbf{-}\mathbf{0}\\ \mathbf{}\mathbf{=}\frac{\mathbf{8}\mathbf{+}\mathbf{7}}{\mathbf{30}}\\ \mathbf{=}\frac{\mathbf{15}}{\mathbf{30}}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\end{array}$

$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}$is the probability that the museum utilizes both total visitors and funds raised as performance measures cannot be done continuously.

Hence, $\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{0}$.

As the tree diagram demonstrates, the probability is as follows:

Because the two museums can utilize any performance measure, they are completely independent. Therefore, the chances of either museum selecting any of the five measures remain the same.

Hence, the sample points' probabilities are as follows:

$\begin{array}{l}\mathbf{Museum}\mathbf{1}\mathbf{.}\\ \mathbf{Total}\mathbf{}\mathbf{visitors}\mathbf{=}\frac{\mathbf{8}}{\mathbf{30}}\mathbf{x}\frac{\mathbf{8}}{\mathbf{30}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{64}}{\mathbf{900}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{16}}{\mathbf{225}}\end{array}$

$\begin{array}{l}\mathbf{Fundvisitors}\mathbf{=}\frac{\mathbf{8}}{\mathbf{30}}\mathbf{x}\frac{\mathbf{7}}{\mathbf{30}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{56}}{\mathbf{900}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{28}}{\mathbf{450}}\end{array}$

$\begin{array}{l}\mathbf{Museum}\mathbf{2}\mathbf{.}\\ \mathbf{Totalvisitors}\mathbf{=}\frac{\mathbf{9}}{\mathbf{30}}\mathbf{x}\frac{\mathbf{8}}{\mathbf{30}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{72}}{\mathbf{900}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{14}}{\mathbf{225}}\end{array}$

$\begin{array}{l}\mathbf{Fund}\mathbf{}\mathbf{visitors}\mathbf{=}\frac{\mathbf{7}}{\mathbf{30}}\mathbf{x}\frac{\mathbf{7}}{\mathbf{30}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{64}}{\mathbf{900}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{49}}{\mathbf{900}}\end{array}$

Hence, the probabilities of museums 1 & 2 are 16/225, 28/450, 14/225, and 49/900.

$$\begin{gathered} \mathbf{P}\mathbf{=}\frac{\mathbf{16}}{\mathbf{225}}\mathbf{+}\frac{\mathbf{28}}{\mathbf{450}}\mathbf{+}\frac{\mathbf{14}}{\mathbf{225}}\mathbf{+}\frac{\mathbf{49}}{\mathbf{900}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{64}\mathbf{+}\mathbf{56}\mathbf{+}\mathbf{56}\mathbf{+}\mathbf{49}}{\mathbf{900}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{225}}{\mathbf{900}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}} \end{gathered}$$

Hence, the probability of both the museum is 1/4.