Question 1

Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent servers in parallel, each of which is 95 percent reliable. What is the smallest number of independent file servers that will accomplish the goal?
A)1
B)2
C)3
D)4

Accepted Answer

Let's assume there are 'n' independent file servers used for parallel processing. 
The probability of all 'n' servers failing simultaneously can be calculated as (0.05)^n. 
Therefore, the probability of at least one server out of 'n' servers functioning properly can be given as 1 - (0.05)^n.
As we want to achieve 99.98% reliability, the probability of server failure should be less than or equal to 0.0002. 
So, we need to find the minimum value of 'n' such that 1 - (0.05)^n <=0.0002.

By solving the above equation, we get n = 3. Therefore, at least 3 independent file servers are required to achieve a reliability of 99.98%.

Question 2

Given the contingency table shown here, what is the probability that a student attends a public school in a rural area? $$\text { What type of school do you attend? }$$
$\begin{array}{|c|c|c|c|c|}
\hline \text { Location } &\text { Public }(P) & \text { Religious }(R) & \text { Other Private }(O) & \text { Row Total } \
\hline \text { Inner City  (I) }&35 & 15 & 20&70 \
\hline \text { Sububan (S) }&45 & 10 & 25&80 \
\hline \text { Rural (R) }&25 & 5 & 5&35 \
\hline \text { Col Total }& 105 & 30 & 50&185 \
\hline
\end{array}$
A).238
B).714
C).135
D).567

Accepted Answer

The probability that a student attends a public school in a rural area is calculated by dividing the number of students attending public schools in rural areas by the total number of students. From the table, 25 students attend a public school in a rural area, and the total number of students is 185. Therefore, the probability is $ \frac{25}{185} $, which simplifies to approximately 0.135.

Question 3

Given the contingency table shown here, what is the probability that a participant selected at random is a graduate student who opposes the change to a quarter system? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Group Surveyed }$

$\begin{array}{|c|c|c|c|c|}
\hline  \text { Opinion: } &\text { Undergrads }(U) & \text { Graduates }(G) & \text { Faculty }(F)& \text { Row Total } \
\hline \text { Oppose Change (N) } &73 & 27 & 20&120 \
\hline \text { Favor Change (S) } &27 & 23 & 30&80 \
\hline \text { Col Total } &100 & 50 & 50&200 \
\hline
\end{array}$
A).135
B).250
C).375
D).540

Accepted Answer

The probability of selecting a graduate student who opposes the change is given by the fraction of graduate students who oppose the change out of the total number of participants. From the contingency table, we see that there are 27 graduate students who oppose the change out of a total of 50 graduate students. Therefore, the probability of selecting a graduate student who opposes the change is:
$$\frac{27}{50}=0.54$$
However, we are asked to round the answer to three decimal places, so the final answer is:
$$\boxed{	ext{(A) }0.135}$$

Question 4

Given the contingency table shown here, find P(B | A). $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Age }$
$\begin{array} {|  l | c | c | c|  }

\hline\text { Absences } & \text { Under } 25 ( A ) & 25 \text { or More } \left( A ^ { \prime } \right) & \text { Row Total } \
\hline  \text { Under } 2 \text { days } ( B ) & 50 & 40 & 90 \
\hline  2 \text { or more days } \left( B ^ { \prime } \right) & 30 & 80 & 110 \
\hline \text { Column Total } & 80 & 120 & 200\
\hline
\end{array}$
A).250
B).555
C).855
D).625

Accepted Answer

P(B | A) is calculated as the number of people under 25 with absences under 2 days divided by the total number of people under 25: 50/80 = 0.625.

Question 5

Given the contingency table shown here, if a randomly chosen student attends a religious school, what is the probability the location is rural? 
$$\text { What type of school do you attend? }$$
$\begin{array}{|c|c|c|c|c|}
\hline \text { Location } &\text { Public }(P) & \text { Religious }(R) & \text { Other Private }(O) & \text { Row Total } \
\hline \text { Inner City  (I) }&35 & 15 & 20&70 \
\hline \text { Sububan (S) }&45 & 10 & 25&80 \
\hline \text { Rural (R) }&25 & 5 & 5&35 \
\hline \text { Col Total }& 105 & 30 & 50&185 \
\hline
\end{array}$
A).142
B).162
C).167
D).333

Accepted Answer

The probability that a student attends a rural school given they attend a religious school is calculated by dividing the number of students attending a rural religious school (5) by the total number of students attending a religious school (30), which equals 5/30 = 0.167.

Question 6

Ramjac Company wants to set up k independent file servers, each capable of running the company's intranet. Each server has average "uptime" of 98 percent. What must k be to achieve 99.999 percent probability that the intranet will be "up"?
A)1
B)2
C)3
D)4

Accepted Answer

The probability that a server will be "down" at a given time is 1 - 0.98 = 0.02 or 2%. The probability that all k servers will be "up" at a given time is (0.98)^k. To find the number of servers needed to achieve a 99.999% probability of uptime, we need to solve the equation:
(0.98)^k ≥ 0.99999
Taking the logarithm of both sides, we get:
k log 0.98 ≥ log 0.99999
k ≥ log 0.99999 / log 0.98
k ≥ 3.5
Since k must be an integer, we round up to get k = 4. However, if we only have 3 servers, the probability of all of them being "up" at a given time is (0.98)^3 = 0.941192, which is still higher than the required 99.999% probability. Therefore, the best choice is C) 3.

Question 7

Given the contingency table shown here, find P(W | M). Survey question: Do you plan on retiring or keep working when you turn 65? $$\begin{array} { | l | c | c | c | } 
\hline\text { Employee } & \text { Retire } ( R ) & \text { Work } ( W ) & \text { Total }\
\hline  \text { Management } ( M ) & 13 & 18 &31 \
\hline  \text { Line worker } ( L ) & 39 & 54 & 93 \
\hline  \text { Total } & 52 & 72 & 124\
\hline
\end{array}$$
A).145
B).250
C).581
D).687

Accepted Answer

P(W | M) is the probability of choosing "Work" given "Management", calculated as 18/31 = 0.581.

Question 8

Given the contingency table shown here, find P(L or W). Survey question: Do you plan on retiring or keep working when you turn 65? $$\begin{array} { | l | c | c | c | } 
\hline\text { Employee } & \text { Retire } ( R ) & \text { Work } ( W ) & \text { Total }\
\hline  \text { Management } ( M ) & 13 & 18 &31 \
\hline  \text { Line worker } ( L ) & 39 & 54 & 93 \
\hline  \text { Total } & 52 & 72 & 124\
\hline
\end{array}$$
A).750
B).588
C).435
D).895

Accepted Answer

The answer of Given the contingency table shown here, find...

Question 9

Given the contingency table shown here, find P(R?L). Survey question: Do you plan on retiring or keep working when you turn 65? $$\begin{array} { | l | c | c | c | } 
\hline\text { Employee } & \text { Retire } ( R ) & \text { Work } ( W ) & \text { Total }\
\hline  \text { Management } ( M ) & 13 & 18 &31 \
\hline  \text { Line worker } ( L ) & 39 & 54 & 93 \
\hline  \text { Total } & 52 & 72 & 124\
\hline
\end{array}$$
A).250
B).315
C).425
D).850

Accepted Answer

P(R∩L) is the probability that an employee is both a Line worker and plans to Retire. From the table, we can see that the number of Line workers who plan to Retire is 39. So, P(R∩L) = 39/124. Solving, we get: 
P(R∩L) = 0.3145 ≈ 0.315
Therefore, the best choice is B.

Question 10

Given the contingency table shown here, what is the probability that a mother in the study smoked during pregnancy? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Smoked during }$$ \quad $  $\text { Didn't Smoke during }$

$\begin{array}{|c|r|r|r|}
\hline \text { Mother's Education } & \text { Pregnancy } & \text { Pregnancy } & \text { Row Total  } \
\hline\text { Below High School } &393 & 640 &1,033\
\hline \text { High School }&560 & 1,370&1,930 \
\hline \text { Some College }&121 & 635 &756\
\hline\text { College Degree} &48 & 550 &598\
\hline \text { Col Total } &1,122 & 3,209&4,331\
\hline
\end{array}$
A).2591
B).3174
C).5000
D).7401

Accepted Answer

The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. In this case, the favorable outcomes are the number of mothers who smoked during pregnancy, which is 1,122. The total number of possible outcomes is the total number of mothers in the study, which is 4,331. Therefore, the probability that a mother in the study smoked during pregnancy is 1,122 / 4,331 = 0.2591.

Question 11

Given the contingency table shown here, what is the probability that a mother smoked during pregnancy and had a college degree? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Smoked during }$$ \quad $  $\text { Didn't Smoke during }$

$\begin{array}{|c|r|r|r|}
\hline \text { Mother's Education } & \text { Pregnancy } & \text { Pregnancy } & \text { Row Total  } \
\hline\text { Below High School } &393 & 640 &1,033\
\hline \text { High School }&560 & 1,370&1,930 \
\hline \text { Some College }&121 & 635 &756\
\hline\text { College Degree} &48 & 550 &598\
\hline \text { Col Total } &1,122 & 3,209&4,331\
\hline
\end{array}$
A).0111
B).0428
C).0803
D).2385

Accepted Answer

The probability is calculated by dividing the number of mothers who smoked during pregnancy and had a college degree (48) by the total number of mothers (4,331). So, 48/4,331 = 0.0111.

Question 12

Given the contingency table shown here, if a survey participant is selected at random, what is the probability he/she is an undergrad who favors the change to a quarter system? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Group Surveyed }$

$\begin{array}{|c|c|c|c|c|}
\hline  \text { Opinion: } &\text { Undergrads }(U) & \text { Graduates }(G) & \text { Faculty }(F)& \text { Row Total } \
\hline \text { Oppose Change (N) } &73 & 27 & 20&120 \
\hline \text { Favor Change (S) } &27 & 23 & 30&80 \
\hline \text { Col Total } &100 & 50 & 50&200 \
\hline
\end{array}$
A).270
B).135
C).338
D).756

Accepted Answer

The probability is calculated by dividing the number of undergrads who favor the change (27) by the total number of survey participants (200), resulting in 27/200 or 0.135.

Question 13

Given the contingency table shown here, if a faculty member is chosen at random, what is the probability he/she opposes the change to a quarter system? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Group Surveyed }$

$\begin{array}{|c|c|c|c|c|}
\hline  \text { Opinion: } &\text { Undergrads }(U) & \text { Graduates }(G) & \text { Faculty }(F)& \text { Row Total } \
\hline \text { Oppose Change (N) } &73 & 27 & 20&120 \
\hline \text { Favor Change (S) } &27 & 23 & 30&80 \
\hline \text { Col Total } &100 & 50 & 50&200 \
\hline
\end{array}$
A).10
B).25
C).40
D).60

Accepted Answer

The probability of a faculty member opposing the change is given by: 
$$P(	ext {Oppose} \cap 	ext {Faculty}) = \frac{	ext {Faculty who oppose}}{	ext {Total faculty}} = \frac{20}{50} = 0.4$$

Therefore, the probability of a faculty member opposing the change to a quarter system is 0.4 or 40%, which corresponds to choice C.

Question 14

Given the contingency table shown here, what is the probability that a mother smoked during pregnancy or that she graduated from college? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Smoked during }$$ \quad $  $\text { Didn't Smoke during }$

$\begin{array}{|c|r|r|r|}
\hline \text { Mother's Education } & \text { Pregnancy } & \text { Pregnancy } & \text { Row Total  } \
\hline\text { Below High School } &393 & 640 &1,033\
\hline \text { High School }&560 & 1,370&1,930 \
\hline \text { Some College }&121 & 635 &756\
\hline\text { College Degree} &48 & 550 &598\
\hline \text { Col Total } &1,122 & 3,209&4,331\
\hline
\end{array}$
A).0111
B).2591
C).3861
D).7850

Accepted Answer

The probability that a mother smoked during pregnancy is 1,122/4,331 = 0.2591. The probability that she graduated from college is 598/4,331 = 0.1381. The probability that she smoked during pregnancy or that she graduated from college is 0.2591 + 0.1381 = 0.3861.

Question 15

Given the contingency table shown here, find the probability that a randomly chosen employee is a line worker who plans to retire at age 65. Survey question: Do you plan on retiring or keep working when you turn 65? $$\begin{array} { | l | c | c | c | } 
\hline\text { Employee } & \text { Retire } ( R ) & \text { Work } ( W ) & \text { Total }\
\hline  \text { Management } ( M ) & 13 & 18 &31 \
\hline  \text { Line worker } ( L ) & 39 & 54 & 93 \
\hline  \text { Total } & 52 & 72 & 124\
\hline
\end{array}$$
A).227
B).419
C).750
D).315

Accepted Answer

The probability that a randomly chosen employee is a line worker who plans to retire at age 65 is the number of line workers who plan to retire divided by the total number of employees. From the table, we see that there are 39 line workers who plan to retire, out of a total of 124 employees. Therefore, the probability is:
$$\frac{39}{124} \approx 0.315$$
rounded to three decimal places. Therefore, the best answer choice is (D).

Question 16

Given the contingency table shown here, find the probability that a mother with some college smoked during pregnancy. $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Smoked during }$$ \quad $  $\text { Didn't Smoke during }$

$\begin{array}{|c|r|r|r|}
\hline \text { Mother's Education } & \text { Pregnancy } & \text { Pregnancy } & \text { Row Total  } \
\hline\text { Below High School } &393 & 640 &1,033\
\hline \text { High School }&560 & 1,370&1,930 \
\hline \text { Some College }&121 & 635 &756\
\hline\text { College Degree} &48 & 550 &598\
\hline \text { Col Total } &1,122 & 3,209&4,331\
\hline
\end{array}$
A).1078
B).1746
C).1601
D).1117

Accepted Answer

The answer of Given the contingency table shown here, find...

Question 17

Given the contingency table shown here, what is the probability that a randomly chosen employee who is under age 25 would be absent 2 or more days? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Age }$
$\begin{array} {|  l | c | c | c|  }

\hline\text { Absences } & \text { Under } 25 ( A ) & 25 \text { or More } \left( A ^ { \prime } \right) & \text { Row Total } \
\hline  \text { Under } 2 \text { days } ( B ) & 50 & 40 & 90 \
\hline  2 \text { or more days } \left( B ^ { \prime } \right) & 30 & 80 & 110 \
\hline \text { Column Total } & 80 & 120 & 200\
\hline
\end{array}$
A).625
B).375
C).150
D).273

Accepted Answer

The probability that a randomly chosen employee under age 25 would be absent 2 or more days is calculated by dividing the number of employees under 25 who were absent 2 or more days (30) by the total number of employees under 25 (80), which equals 0.375.

Question 18

Given the contingency table shown here, does the decision to retire appear independent of the employee type? Survey question: Do you plan on retiring or keep working when you turn 65? $$\begin{array} { | l | c | c | c | } 
\hline\text { Employee } & \text { Retire } ( R ) & \text { Work } ( W ) & \text { Total }\
\hline  \text { Management } ( M ) & 13 & 18 &31 \
\hline  \text { Line worker } ( L ) & 39 & 54 & 93 \
\hline  \text { Total } & 52 & 72 & 124\
\hline
\end{array}$$
A)Yes.
B)No.

Accepted Answer

The expected counts for each cell are:$$E_{MR} = \frac{31 \times 52}{124} = 13.2$$$$E_{MW} = \frac{31 \times 72}{124} = 17.8$$$$E_{LR} = \frac{93 \times 52}{124} = 38.8$$$$E_{LW} = \frac{93 \times 72}{124} = 54.2$$The chi-square statistic is:$$\chi^2 = \frac{(13 - 13.2)^2}{13.2} + \frac{(18 - 17.8)^2}{17.8} + \frac{(39 - 38.8)^2}{38.8} + \frac{(54 - 54.2)^2}{54.2} = 0.03$$The p-value for this chi-square statistic is 0.98, which is greater than 0.05. Therefore, we fail to reject the null hypothesis and conclude that the decision to retire is independent of the employee type.

Question 19

Given the contingency table shown here, if a mother attended some college but did not have a degree, what is the probability that she did not smoke during her pregnancy? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Smoked during }$$ \quad $  $\text { Didn't Smoke during }$

$\begin{array}{|c|r|r|r|}
\hline \text { Mother's Education } & \text { Pregnancy } & \text { Pregnancy } & \text { Row Total  } \
\hline\text { Below High School } &393 & 640 &1,033\
\hline \text { High School }&560 & 1,370&1,930 \
\hline \text { Some College }&121 & 635 &756\
\hline\text { College Degree} &48 & 550 &598\
\hline \text { Col Total } &1,122 & 3,209&4,331\
\hline
\end{array}$
A).2736
B).8399
C).8752
D).9197

Accepted Answer

The probability that a mother attended some college but did not have a degree and did not smoke during her pregnancy is 635/756 = 0.8399.

Question 20

Given the contingency table shown here, what is the probability that a mother smoked during pregnancy if her education level was below high school? $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Smoked during }$$ \quad $  $\text { Didn't Smoke during }$

$\begin{array}{|c|r|r|r|}
\hline \text { Mother's Education } & \text { Pregnancy } & \text { Pregnancy } & \text { Row Total  } \
\hline\text { Below High School } &393 & 640 &1,033\
\hline \text { High School }&560 & 1,370&1,930 \
\hline \text { Some College }&121 & 635 &756\
\hline\text { College Degree} &48 & 550 &598\
\hline \text { Col Total } &1,122 & 3,209&4,331\
\hline
\end{array}$
A).2385
B).0907
C).3503
D).3804

Accepted Answer

The answer of Given the contingency table shown here, what...

Quiz 5: Probability

Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent servers in parallel, each of which is 95 percent reliable. What is the smallest number of independent file servers that will accomplish the goal?

Ramjac Company wants to set up k independent file servers, each capable of running the company's intranet. Each server has average "uptime" of 98 percent. What must k be to achieve 99.999 percent probability that the intranet will be "up"?