Question 1

In any sample space P(A | B) and P(B | A):
A)are always equal to one another.
B)are never equal to one another.
C)are reciprocals of one another.
D)are equal only if P(A) = P(B).

Accepted Answer

P(A | B) and P(B | A) are equal only if P(A) = P(B). This is because P(A | B) = P(A ∩ B) / P(B) and P(B | A) = P(B ∩ A) / P(A), and for these to be equal, P(A) must equal P(B).

Question 2

If a call is sampled at random, the conditional probability that the call is not "6+" minutes long given that it is a long distance call is: $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Call Length (minutes) }$
$\begin{array}{|c|c|r|c|c|}
\hline \text {Type of Phone Call} & 0-1 & 2-5 & 6+ & \text { Total } \
\hline \text { Local } &150 & 250 & 100 & 500 \
\hline \text { Long Distance } &170 & 120 & 10 \
\hline \text { Total } &320 & 370 & 110 & 300 \
\hline
\end{array}$
A)120/300.
B)10/300.
C).9667.
D).6667.

Accepted Answer

The total number of long distance calls is 300 (170 + 120 + 10). The number of long distance calls that are not "6+" minutes long is 290 (170 + 120). Therefore, the conditional probability is 290/300 = .9667.

Question 3

Given the contingency table shown here, find the joint probability that a call sampled at random out of this population is local and 2-5 minutes long. $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Call Length (minutes) }$
$\begin{array}{|c|c|r|c|c|}
\hline \text {Type of Phone Call} & 0-1 & 2-5 & 6+ & \text { Total } \
\hline \text { Local } &150 & 250 & 100 & 500 \
\hline \text { Long Distance } &170 & 120 & 10 \
\hline \text { Total } &320 & 370 & 110 & 300 \
\hline
\end{array}$
A).5000
B).3125
C).4000
D).4625

Accepted Answer

The joint probability of a call being local and 2-5 minutes long is the intersection of the Local row and the 2-5 column divided by the total number of calls. This is given by:

(250/500) / (880/300) = 3125/880

Simplifying this fraction gives 25/8, which is approximately 3.125. Therefore, the answer is B).

Question 4

Debbie has two stocks, X and Y. Consider the following events: X = the event that the price of stock X has increased
Y = the event that the price of stock Y has increased
The event "the price of stock X has increased and the price of stock Y has not increased" may be written as
A)X′∩Y
B)X or Y′
C)X∩Y′
D)X or Y

Accepted Answer

The event "the price of stock X has increased and the price of stock Y has not increased" means that X has occurred but Y has not occurred. This can be represented as the intersection of X and the complement of Y (denoted by Y′). Therefore, the correct choice is C, which is X∩Y′.

Question 5

Given the contingency table shown here, find the probability that a randomly chosen individual is a female economics major. $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Major }$
$\begin{array} {|  c | c | c | c | c|  } 
\hline\text { Gender } & \text { Accounting } ( A ) & \text { Gen. Mgmt. } ( G ) & \text { Economics } ( E ) & \text { Row Total } \
\hline\text { Male } ( M ) & 210 & 180 & 140 & 530 \
\hline  \text { Female } ( F ) & 150 & 160 & 160 & 470 \
\hline  \quad \text { Col Total } & 360 & 340 & 300 & 1000\
\hline
\end{array}$
A).3404
B).4700
C).1600
D).5333

Accepted Answer

To find the probability that a randomly chosen individual is a female economics major, we need to look at the cell in the contingency table that corresponds to this category, which has a value of 160. To get the probability, we divide this value by the total number of individuals in the sample, which is 1000. Therefore, the probability is 160/1000 = 0.16 or 16%. The answer that matches this is choice C).

Question 6

If P(A) = 0.35, P(B) = 0.60, and P(A or B) = 0.70, then:
A)A and B are mutually exclusive.
B)P(A∩B) = .15.
C)P(A∩B) = .25.
D)P(A∩B) = .35.

Accepted Answer

Since P(A or B) = P(A) + P(B) - P(A∩B), we can plug in the given values to get:
0.70 = 0.35 + 0.60 - P(A∩B)
0.15 = P(A∩B)
Therefore, the answer is (C) P(A∩B) = 0.25.

Question 7

The following probabilities are given about events A and B in a sample space: P(A) = 0.30, P(B) = 0.40, P(A or B) = 0.60. We can say that:
A)P(A∩B) = 0.70.
B)P(A) = P(A∩B).
C)P(A∩B) = 0.10.
D)A and B are independent events.

Accepted Answer

The probability of A or B can be calculated as P(A or B) = P(A) + P(B) - P(A∩B). Plugging in the given values, we have 0.6 = 0.3 + 0.4 - P(A∩B), which gives us P(A∩B) = 0.1. Therefore, option C is correct. Option A is incorrect because it suggests that P(A∩B) = 0.7, which would mean that P(A or B) > 1, violating the basic rules of probability. Option B is incorrect because it suggests that P(A) = P(A∩B), which is not generally true unless A and B are mutually exclusive events. Option D cannot be determined from the information given.

Question 8

If P(A | B) = 0.40 and P(B) = 0.30, find P(A∩B).
A).171
B).525
C).571
D).120

Accepted Answer

P(A∩B) = P(A | B) * P(B) = 0.40 * 0.30 = 0.12.

Question 9

The following relationship always holds true for events A and B in a sample space.
A)P(A | B) = P(B | A)
B)P(A∩B) = P(A | B) P(B)
C)P(A | B) = P(B | A) P(A)

Accepted Answer

The formula P(A∩B) = P(A | B) P(B) correctly describes the relationship between the joint probability of events A and B occurring together and the conditional probability of A given B multiplied by the probability of B. This is a fundamental principle in probability theory.

Question 10

If P(A∩B) = 0.50, can P(A) = 0.20?
A)Only if P(A | B) = 0.10
B)Not unless P(B) = 0.30
C)Only if P(B∩A) = 0.60
D)If P(A) = 0.20, then P(A∩B) cannot equal 0.50.

Accepted Answer

The probability of the intersection of A and B (P(A∩B)) cannot be greater than the probability of either event individually. If P(A) = 0.20, then P(A∩B) cannot be 0.50, as the intersection cannot have a higher probability than either event alone.

Question 11

The following table shows the survival experience of 1,000 males who retire at age 65: $$\begin{array} { c c } 
\text { Age } & \text { Number of Males Surviving } \
65 & 1,000 \
70 & 907 \
75 & 775 \
80 & 596 \
85 & 383
\end{array}$$ Based on these data, the probability that a 75-year-old male will survive to age 80 is:
A)0.596
B)1 - 0.596 = 0.404
C)1 - 0.775 = 0.225
D)0.769

Accepted Answer

The probability that a 75-year-old male will survive to age 80 is calculated by dividing the number of males surviving to age 80 (596) by the number of males surviving to age 75 (775), which gives 0.769.

Question 12

Given the contingency table shown here, find P(A?M). $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Major }$
$\begin{array} {|  c | c | c | c | c|  } 
\hline\text { Gender } & \text { Accounting } ( A ) & \text { Gen. Mgmt. } ( G ) & \text { Economics } ( E ) & \text { Row Total } \
\hline\text { Male } ( M ) & 210 & 180 & 140 & 530 \
\hline  \text { Female } ( F ) & 150 & 160 & 160 & 470 \
\hline  \quad \text { Col Total } & 360 & 340 & 300 & 1000\
\hline
\end{array}$
A).210
B).360
C).396
D).583

Accepted Answer

P(A∩M) refers to the probability of being both in the Accounting major and Male, which is directly given as 210 in the table.

Question 13

Given the contingency table shown here, if a call is sampled at random, find the marginal probability that the call is long distance. $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Call Length (minutes) }$
$\begin{array}{|c|c|r|c|c|}
\hline \text {Type of Phone Call} & 0-1 & 2-5 & 6+ & \text { Total } \
\hline \text { Local } &150 & 250 & 100 & 500 \
\hline \text { Long Distance } &170 & 120 & 10 \
\hline \text { Total } &320 & 370 & 110 & 300 \
\hline
\end{array}$
A).3750
B).6250
C).4000
D)300/500

Accepted Answer

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

Question 14

The following table gives a classification of the 10,000 shareholders of Oxnard Xylophone Distributors, Inc. A few numbers are missing from the table. Given that a shareholder holding 500-999 shares is picked, there is a 0.625 probability that the shareholder will be a woman. Consequently, what is the number of men holding 1000 or more shares? $\begin{array}{|r|r|r|r|r|}
\hline \text { Shareholders } &{-499} & 500-999 & 1,000+ &\text { Total }\
\hline \text { Men } &3,000 & & &4,000\
\hline \text { Women } &2,000 & & &4,800\
\hline \text { Joint Accounts} &0 & 1,000 & 200 &1,200\
\hline \text { Total } &5,000 & 4,000 & 1,000& 10,000  \
\hline
\end{array}$
A)1,000
B)250
C)7,500
D)500

Accepted Answer

Let the number of women holding 500-999 shares be $w$. Then, we know that $\frac{w}{w+(4000-m)+200}=0.625$, where $m$ is the number of men holding 1000 or more shares. Simplifying this equation gives us $w=10m-2000$.

We also know that the total number of shareholders holding 500-999 shares is $4000-3000-1000=0$, so $w=0$. Therefore, $m=200$ and the answer is $\boxed{	extbf{(D)}\ 500}$.

Question 15

A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of typeB. Five percent of all type A goggles are returned within 10 days after the sale, whereas only two percent of type B are returned. If a pair of goggles is returned within the first 10 days after the sale, the probability that the goggles returned are of type B is:
A).014.
B).140.
C).070.
D).483.

Accepted Answer

The probability that a returned pair of goggles is of type B can be calculated using Bayes' theorem. Let's denote $P(A)$ as the probability of producing type A goggles, $P(B)$ as the probability of producing type B goggles, $P(R|A)$ as the probability of type A goggles being returned, and $P(R|B)$ as the probability of type B goggles being returned. We want to find $P(B|R)$, the probability that a returned pair of goggles is of type B.Given:- $P(A) = 0.30$- $P(B) = 0.70$ (since 100% - 30% = 70%)- $P(R|A) = 0.05$- $P(R|B) = 0.02$The probability that a pair of goggles is returned, $P(R)$, can be calculated as:$$P(R) = P(R|A)P(A) + P(R|B)P(B) = 0.05 \times 0.30 + 0.02 \times 0.70 = 0.029$$Using Bayes' theorem, the probability that a returned pair of goggles is of type B, $P(B|R)$, is:$$P(B|R) = \frac{P(R|B)P(B)}{P(R)} = \frac{0.02 \times 0.70}{0.029} \approx 0.483$$Therefore, the correct answer is D, which is approximately 0.483.

Question 16

Given the contingency table shown here, find P(E). $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Major }$
$\begin{array} {|  c | c | c | c | c|  } 
\hline\text { Gender } & \text { Accounting } ( A ) & \text { Gen. Mgmt. } ( G ) & \text { Economics } ( E ) & \text { Row Total } \
\hline\text { Male } ( M ) & 210 & 180 & 140 & 530 \
\hline  \text { Female } ( F ) & 150 & 160 & 160 & 470 \
\hline  \quad \text { Col Total } & 360 & 340 & 300 & 1000\
\hline
\end{array}$
A).180
B).300
C).529
D).641

Accepted Answer

P(E) refers to the probability of selecting someone who is majoring in Economics, which is the total number of students majoring in Economics (300) divided by the total number of students (1000), hence P(E) = 300/1000 = 0.3. The question asks for the number of students, not the probability, so the correct answer is the total number of students majoring in Economics, which is 300.

Question 17

Given the contingency table shown here, if a randomly chosen student attends school in an inner-city location, what is the probability that it is a public school? $$\text { What type of school do you attend? }$$
$\begin{array}{|c|c|c|c|c|}
\hline \text { Location } &\text { Public }(P) & \text { Religious }(R) & \text { Other Private }(O) & \text { Row Total } \
\hline \text { Inner City  (I) }&35 & 15 & 20&70 \
\hline \text { Sububan (S) }&45 & 10 & 25&80 \
\hline \text { Rural (R) }&25 & 5 & 5&35 \
\hline \text { Col Total }& 105 & 30 & 50&185 \
\hline
\end{array}$
A).189
B).333
C).500
D).567

Accepted Answer

The probability of a student attending a public school given that they attend school in the inner city is found by taking the number of inner city students who attend public schools (35) and dividing by the total number of inner city students (70):

$P(	ext{Public at Inner City}) = \frac{35}{70} = 0.5$

So the answer is C) 0.5.

Question 18

Given the contingency table shown here, find P(F or G). $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Major }$
$\begin{array} {|  c | c | c | c | c|  } 
\hline\text { Gender } & \text { Accounting } ( A ) & \text { Gen. Mgmt. } ( G ) & \text { Economics } ( E ) & \text { Row Total } \
\hline\text { Male } ( M ) & 210 & 180 & 140 & 530 \
\hline  \text { Female } ( F ) & 150 & 160 & 160 & 470 \
\hline  \quad \text { Col Total } & 360 & 340 & 300 & 1000\
\hline
\end{array}$
A).160
B).470
C).650
D).810

Accepted Answer

P(F or G) = P(F) + P(G) - P(F and G)= 470/1000 + 340/1000 - 160/1000= 650/1000= 0.65

Question 19

Given the contingency table shown here, find P(G | M). $\begin{array}{|c|r|r|r|r|}
\hline \begin{array}{c}
\text { Vehicle Type} \

\end{array} & \begin{array}{c}
\text { Somerset } \
(S)
\end{array} & \begin{array}{r}
\text { Oakland } \
(O)
\end{array} & \begin{array}{r}
\text { Great Lakes } \
(G)
\end{array} & \begin{array}{r}
\text { Row Total } \

\end{array} \

\hline  \text { Car (C) }&44 & 49 & 36&129 \
\hline \text { Minivan  (M) }& 21 & 15 & 18&54 \
\hline \text { Full-Size Van (F)  }&2 & 3 & 3&8 \
\hline \text { SUV (I) }&19 & 27 & 26 &72\
\hline \text { Truck (T) }&14 & 6 & 17 &37\
\hline \text { Col Total}&100 & 100 & 100 &400\
\hline
\end{array}$
A).1800
B).0450
C).3333
D).1350

Accepted Answer

P(G | M) means the probability of a vehicle being Great Lakes given that it is a Minivan.

From the table, we see that there are a total of 54 minivans, out of which 18 are from Great Lakes.

So, P(G | M) = 18/54 = 0.3333.

Therefore, the answer is C.

Question 20

Given the contingency table shown here, find P(E | F). $ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$ \quad $$\text { Major }$
$\begin{array} {|  c | c | c | c | c|  } 
\hline\text { Gender } & \text { Accounting } ( A ) & \text { Gen. Mgmt. } ( G ) & \text { Economics } ( E ) & \text { Row Total } \
\hline\text { Male } ( M ) & 210 & 180 & 140 & 530 \
\hline  \text { Female } ( F ) & 150 & 160 & 160 & 470 \
\hline  \quad \text { Col Total } & 360 & 340 & 300 & 1000\
\hline
\end{array}$
A).160
B).300
C).340
D).533

Accepted Answer

P(E | F) represents the probability of being in Economics given that the person is Female. From the table, there are 160 females in Economics out of a total of 470 females, so P(E | F) = 160/470. The correct choice is based on understanding conditional probability, not the raw numbers themselves, hence none of the provided choices directly represent the correct probability value.

Quiz 5: Probability

In any sample space P(A | B) and P(B | A) :

Debbie has two stocks, X and Y. Consider the following events: X = the event that the price of stock X has increased Y = the event that the price of stock Y has increased The event "the price of stock X has increased and the price of stock Y has not increased" may be written as

If P(A) = 0.35, P(B) = 0.60, and P(A or B) = 0.70, then:

The following probabilities are given about events A and B in a sample space: P(A) = 0.30, P(B) = 0.40, P(A or B) = 0.60. We can say that:

If P(A | B) = 0.40 and P(B) = 0.30, find P(A∩B) .

The following relationship always holds true for events A and B in a sample space.

If P(A∩B) = 0.50, can P(A) = 0.20?