QUIZ 1

Exersice 1

Thirty percent of the people on the gulf coast think that the federal government’s response to hurricane Katrina was acceptable. Fifty five percent of the people on the gulf coast are registered republicans. Fur- thermore, forty percent of the republicans think that the response was acceptable.

1. What percent of the people are not republicans and feel that the response was acceptable?

Let´s denote:

S: people of the golf coast

R: percent of the people on the gulf coast are registered republicans

A: percent of the people on the gulf coast think that government’s response was acceptable

\(\bold{R\cap A:}\) republicans that think that the federal government´s response was acceptable

then:

therefore:

\[ \bf{R^c\cap A = 8\%}\]

2. What is the probability that someone thinks the response was acceptable?

\[ \bf{P(A)= \frac{|A|}{|S|}}=\frac{30}{100}= 0.3\] c. Given someone is a Republican, what is the probability they feel the response was acceptable?

\[ \bf{P(A|R)= \frac{(A\cap R)}{P(R)}}=\frac{\frac{22}{100}}{\frac{55}{100}}= 0.4\] d. Given someone is not a Republican, what is the probability they feel the response was acceptable?

being \[\bf{P(R^c)}= 1-R = 1-.55 = .45\] \[ \bf{P(A|R^c)= \frac{(A\cap R^C)}{P(R^c)}}=\frac{\frac{8}{100}}{.45}= 0.17\]

5. What to the values in (b), (c) and (d) imply about the relations between attitude toward response and political affiliation.

Th values in b, c and d implies that there is a positive correlation in the people on the gulf coast think that the federal government´s response to hurricane Katrina was acceptable and their political afiliation.

Exersice 2

You are presented with 3 boxes, each of which contains 8 marbles. The boxes have the following combinations of red and white marbles:

You select a marble at random from Box A. If it is red, you select a second marble from Box B; if, however, your first marble is white, then you choose your second marble from Box C. Using R1, W1 and R2, W2 to signify the color of the first and second marbles, respectively:

1. Draw a probability tree for this process.

Under the suppose that the event to take the first ball and the second ball are independent

Then:

2. Find the probability that exactly one red and one white marble is chosen. \[ \bf{P(R)= \frac{|R|}{|A|}}=\frac{6}{8}= 0.75\] \[ \bf{P(B)= \frac{|W|}{|B|}}=\frac{2}{8}= 0.25\] \[ \bf{P(R \cap W)}= 0.75 * 0.25 = 0.187\]
3. Find the probability that either two red or two white marbles are chosen.

being:

\(\bf{A^1}\) the event of having 2 balls red

\(\bf{W^1}\) the event of having 2 balls white

then: \[ \bf{P(R) en A} = 0.75\] \[ \bf{P(R) en B} = 0.75\] \[ \bf{P(R^2)=P(R \cap R)}= 0.75 * 0.75 = 0.562\] \[ \bf{P(W) en A} = 0.25\] \[ \bf{P(W) en C} = 0.50\] \[ \bf{P(W^2)=P(W \cap W)}= 0.25 * 0.50 = 0.125\] \[ \bf{P(R^2 \cup W^2)}= 0.562 + 0.125 - 0 = 0.687\] d. Find the probability that either the first marble chosen is white OR the second marble chosen is red. \[ \bf{P(W \cup R)}= \frac{2}{8} + \frac{44}{64} -\frac{8}{64} = 0.812\]

Exercise 3

The memo below is from then California governor Arnold Schwarzenegger to the California legislative as- sembly. The first letter of each row in the third and fourth paragraph reveal some colorful language. The governor claimed it was a coincidence that his memo happened to spell out his exact feelings toward the legislature.

1. What is the probability of these two words being spelled if a coincidence occurred and if each letter of the alphabet is equally likely?

Let`s denote:

A: the event that these two words being spelled

S: the number of permutations with replacement considering that for these two words to be formed, the beginnings of seven sentences are taken into consideration

There are permutations with replacement because all the letter of the alphabet are consider, the order matters and the elements are repeated

\[\bf {S^{26}_{7}= \frac{26!}{(26-7)!} = 3,315,312,000 } \]

\[\bf {A= 1}\]

\[\bf {P(A)= \frac {|A|}{|S|} = \frac {1}{3,315,312,000} =3.01x10^{-10}} = 0.000000000301 \]

2. What is the probability of any two words being formed at the beginning of the paragraphs, by random chance?

Let´s denote:

|F|: the amount words of four letters in English = 3,996

|A|: the total amount of possibilities to arrange four spaces with the 26 letters of the alphabet \(\bf{=26^{4}} = 456,976\)

|T|: the amount words of three letters in English = 1,065

|B|: the total amount of possibilities to arrange three spaces with the 26 letters of the alphabet \(\bf{=26^{3}} = 17,576\)

assuming independence between F and T

\[\bf {P(F \cap T)= \frac {|F|}{|A|} * \frac {|T|}{|B|} = \frac {3,996}{456,976} * \frac {1,065}{17,576}= 5.2x10^{-07}=0.00000052} \]

3. How would you determine the probability of a colorful epithet being formed at the beginning of these two paragraphs? (you don’t need to calculate this, just walk us through the steps you’d take).

I would try to establish the cardinality of the permutations within the 26 letters of the English alphabet for the 7 spaces to form the words. Then the cardinality to form that words that is only arrange. Then, I would divide the cardinality of the event by the cardinality of the total of permutations.

Exercise 4

Census data shows that 3.2% of Mexican residents are 75 years of age or older, and that 51.2% of Mexican residents are female. Is it therefore correct to conclude that because (0.032)(0.512) = 0.0164, approximately 2% of Mexican residents are females age 75 or above? Explain.

Is correct to conclude that \(0.0164\) of Mexicans residents are females age 75 or above only if we assume independence between the two events.

Let´s denote:

A: Mexican residents are 75 years of age or older

B: Mexican residents are female

Therefore, under independence:

\[ \bf{P(A \cap B)}= P(A) * P(B) = 0.0164\]

Exercise 5

In Mexico, the probability that a live-born infant is male is 0.505. Assume that the sexes of newborns are independent events.

1. Given two infants where it is known that infant 1 is male, what is the probability that infant 2 is male?

Let´s denote:

P(B): the probability that a live-born is male = 0.505

under independence:

\[ \bf{P(B | B)}= P(B) = 0.505\] b. What is the probability that a live-born infant is not male?

\[ \bf{P(B^c)}= 1 -P(B^c) =1 - 0.505 = 0.495\] c. What is the probability that two infants are both male?

\[ \bf{P(B \cap B)}= P(B) * P(B) = 0.505 * 0.505 = .255 \] d. What is the probability that five infants are all male?

\[ \bf{P(B \cap B \cap B \cap B \cap B)}= P(B)^5 = 0.505^5 = 0.032 \] e. What is the probability that five infants are NOT all male?

\[ \bf{P(A \cap A \cap A \cap A \cap A)}= P(A)^5 = 0.495^5 = 0.0297 \]

Exercise 6

Using this information, answer the questions below:

1. Given that the system has a type 1 defect, what is the probability that it has a type 2 defect?

\[ \bf{P(A2|A1)}= \frac{P(A2 \cap A1)}{P(A1)} + \frac{0.03}{0.12} = 0.25\]

2. Given that the system has a type 1 defect, what is the probability that it has all three types of defects? \[ \bf{P(A1 \cap A2 \cap A3 | A1)}= \frac{P(A1 \cap A2 \cap A3)}{P(A1)} + \frac{0.01}{0.12} = 0.083\]

3. Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? \[\begin{equation} \begin{split} \bf{P((A1/A2 A3)\cup (A2/A1A3) \cup (A3/A1A2) | (A1 \cup A2 \cup A3))} \\ & =\frac{P(A1/A2 A3) + P(A2/A1A3) + P(A3/A1A2)}{P(A1 \cup A2 \cup A3)} \\ & =\frac{0.04 +0.01 +0}{0.14} \\ & \bf= 0.357 \end {split} \end{equation}\]

4. Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect? \[ \bf{P(A3^c| A1\cap A2)}= \frac{P(A3^c \cap(A1 \cap A2))}{P(A1 \cap A2)} + \frac{0.02}{0.03} = 0.66\]

Exercise 7

Mexican Senate contains four senators from each of the 32 states.

1. If a committee of eight senators is selected at random, what is the probability that it will contain at least one of the four senators from a certain specified state?

Let´s denote:

\(\bf{|S|:}\) 128

A: at least one for each state

\[\begin{equation} \bf\mathbf {{{P(A)}= 1-\frac{\binom{124}{8}}{\binom{128}{8}} = 1- .77= 0.229}}\end{equation}\]

2. What is the probability that a group of 32 senators selected at random will contain one senator from each state?

being:

B: one senator from each state for the group

\[\begin{equation} \bf\mathbf{{P(B)}= \frac{4^{32}}{\binom{128}{32}}= 1.24e^{-11}} \end{equation}\]

Exercise 8

Star Medica Hospital in Aguascalientes is considering an expansion of its facilities. The proposed enlargement has great support among the hospital’s medical and research staff. The financial office, however, fears that demand for the hospital’s services will not grow enough to warrant the building of new capacity. It has been estimated that there is a 0.8 probability that the new facility will break even if the demand for services rises or remains stable over the next 5 years. On the other hand, there is only a 0.3 probability of breaking even if demand over that period drops. Recent developments on the national health care front suggest a 0.75 probability that demand will fall over the next 5 years. What is the probability that the facility, if built, will break even?

Let´s denote:

B: the hospital will break

A1: demand of hospital´s services will grow

A2: demand of hospital´s services will fall

by Bayes Theorem

\[\begin{equation} \bf {{P(B)}= P(A2)P(B|A2) + P(A1)P(B|A1) = (0.75)(0.3) + (0.25)(0.8) = .225 + .2 = 0.425} \end{equation}\]

Exercise 9

There are two tests available for the diagnosis of Greek Syndrome (GS), a hypothetical disease that causes math phobia. Data from a study of the Alpha test and Beta test by true disease status are shown in the following table. The prevalence of GS in the population of interest – graduate students – is 10%. Based on the tables above, respond the following questions:

1. Calculate the revised probability of GS given a positive result on the Alpha test.

Let´s denote:

C: the event of having GS

\(\bf{{C^c:}}\) the event of not having GS

\(\bf {{A_{+}:}}\) the event of that Alpha Test is positive

\(\bf{{A_{-}:}}\) the event of that Alpha Test is negative

\(\bf{{B_{+}:}}\) the event of that Beta Test is positive

\(\bf{{B_{-}:}}\) the event of that Beta Test is negative

P(C) = 0.1

\(\bf{P(C^c)}\) =0.9

\[\begin{equation} \bf\mathbf{{P(C|A_{+})}= \frac{P(A_{+}|C)P(C)}{P(A_{+})} = \frac {P(A_{+}|C)P(C)}{P(A_{+}|C)P(C) + P(A_{+}|C^c)P(C^c)} = \frac{(0.3)(0.1)}{(0.3)(0.1)+(0.9)(0.1)}= \frac{(0.3)}{(0.12)} = 0.25}\end{equation}\]

2. Calculate the revised probability of GS given a positive result on the Beta test.

\[\begin{equation} \bf\mathbf{{P(C|B_{+})}= \frac{P(B_{+}|C)P(C)}{P(B_{+})} = \frac {P(B_{+}|C)P(C)}{P(B_{+}|C)P(C) + P(B_{+}|C^c)P(C^c)} = \frac{(0.4)(0.1)}{(0.4)(0.1)+(0.05)(0.9)}= \frac{(0.04)}{(0.085)} = 0.47}\end{equation}\]

3. Calculate the revised probability of GS, assuming conditional independence given the presence or absence of disease, for the following:

1. Positive results on both Alpha and Beta tests.

\[\begin{equation} \begin{split} \bf{P(C|A_{+} \cap B_{+})}= \frac{P(A_{+} \cap B_{+}|C)P(C)}{P(A_{+} \cap B_{+})} \\ & = \frac {P(A_{+} \cap B_{+}|C)P(C)}{P(A_{+} \cap B_{+}|C)P(C) + P(A_{+} \cap B_{+}|C^c)P(C^c)} \\ & = \frac{(0.3)(0.4)(0.1)}{(0.3)(0.4)(0.1)+(0.9)(0.1)(0.05)} \\ & = \frac{(0.012)}{(0.016)} \\ & \bf{= 0.75} \end{split} \end{equation}\]

2. A positive result on the Alpha test and a negative result on the Beta test.

\[\begin{equation} \begin{split} \bf{P(C|A_{+} \cap B_{-})}= \frac{P(A_{+} \cap B_{-}|C)P(C)}{P(A_{+} \cap B_{-})} \\ & = \frac {P(A_{+} \cap B_{-}|C)P(C)}{P(A_{+} \cap B_{-}|C)P(C) + P(A_{+} \cap B_{-}|C^c)P(C^c)} \\ & = \frac{(0.3)(0.6)(0.1)}{(0.3)(0.6)(0.1)+(0.1)(0.95)(0.9)} \\ & = \frac{(0.018)}{(0.103)} \\ & \bf{= 0.17} \end{split} \end{equation}\]

3. A positive result on the Beta test and a negative result on the Alpha test.

\[\begin{equation} \begin{split} \bf{P(C|A_{-} \cap B_{+})}= \frac{P(A_{-} \cap B_{+}|C)P(C)}{P(A_{-} \cap B_{+})} \\ & = \frac {P(A_{-} \cap B_{+}|C)P(C)}{P(A_{-} \cap B_{+}|C)P(C) + P(A_{-} \cap B_{+}|C^c)P(C^c)} \\ & = \frac{(0.4)(0.7)(0.1)}{(0.4)(0.7)(0.1)+(0.9)(0.05)(0.9)} \\ & = \frac{(0.028)}{(0.068)} \\ & \bf{= 0.41} \end{split} \end{equation}\]

4. Negative results on both Alpha and Beta tests.

\[\begin{equation} \begin{split} \bf{P(C|A_{-} \cap B_{-})}= \frac{P(A_{-} \cap B_{-}|C)P(C)}{P(A_{-} \cap B_{-})} \\ & = \frac {P(A_{-} \cap B_{-}|C)P(C)}{P(A_{-} \cap B_{-}|C)P(C) + P(A_{-} \cap B_{-}|C^c)P(C^c)} \\ & = \frac{(0.1)(0.6)(0.7)}{(0.1)(0.6)(0.7)+(0.9)(0.9)(0.95)} \\ & = \frac{(0.042)}{(0.811)} \\ & \bf{= 0.51} \end{split} \end{equation}\]

4. Actually, the Alpha and Beta tests are not conditionally independent. In fact, if the result of the Alpha test is positive, it guarantees that the Beta test is positive given the presence of GS. The joint characteristics of the tests are shown in the table below.

What is the probability of disease given:

1. Positive results on both Alpha and Beta tests.

\[\begin{equation} \begin{split} \bf{P(C|A_{+} \cap B_{+})}= \frac{P(A_{+} \cap B_{+}|C)P(C)}{P(A_{+} \cap B_{+})} \\ & = \frac {P(A_{+} \cap B_{+}|C)P(C)}{P(A_{+} \cap B_{+}|C)P(C) + P(A_{+} \cap B_{+}|C^c)P(C^c)} \\ & = \frac{(0.3)(0.1)}{(0.3)(0.1)+(0.9)(0.003)} \\ & = \frac{(0.03)}{(0.0345)} \\ & \bf{= 0.88} \end{split} \end{equation}\]

2. A positive result on the Alpha test and a negative result on the Beta test.

\[\begin{equation} \begin{split} \bf{P(C|A_{+} \cap B_{-})}= \frac{P(A_{+} \cap B_{-}|C)P(C)}{P(A_{+} \cap B_{-})} \\ & = \frac {P(A_{+} \cap B_{-}|C)P(C)}{P(A_{+} \cap B_{-}|C)P(C) + P(A_{+} \cap B_{-}|C^c)P(C^c)} \\ & = \frac{(0.0)(0.1)}{(0.0)(0.1)+(0.1)(0.95)(0.9)} \\ & = \frac{(0.0)}{(0.08)} \\ & \bf{= 0.0} \end{split} \end{equation}\]

3. A positive result on the Beta test and a negative result on the Alpha test.

\[\begin{equation} \begin{split} \bf{P(C|A_{-} \cap B_{+})}= \frac{P(A_{-} \cap B_{+}|C)P(C)}{P(A_{-} \cap B_{+})} \\ & = \frac {P(A_{-} \cap B_{+}|C)P(C)}{P(A_{-} \cap B_{+}|C)P(C) + P(A_{-} \cap B_{+}|C^c)P(C^c)} \\ & = \frac{(0.1)(0.1)}{(0.1)(0.1)+(0.9)(0.045)} \\ & = \frac{(0.01)}{(0.05)} \\ & \bf{= 0.2} \end{split} \end{equation}\]

4. Negative results on both Alpha and Beta tests.

\[\begin{equation} \begin{split} \bf{P(C|A_{-} \cap B_{-})}= \frac{P(A_{-} \cap B_{-}|C)P(C)}{P(A_{-} \cap B_{-})} \\ & = \frac {P(A_{-} \cap B_{-}|C)P(C)}{P(A_{-} \cap B_{-}|C)P(C) + P(A_{-} \cap B_{-}|C^c)P(C^c)} \\ & = \frac{(0.1)(0.6)}{(0.1)(0.6)+(0.9))(0.853)} \\ & = \frac{(0.006)}{(0.775)} \\ & \bf{= 0.007} \end{split} \end{equation}\]

5. How do your answers to questions #3 and #4 compare?

When we don´t assume independence the information given by the test is more precise

Exercise 10

Two litters* of a particular rodent species have been born, one with two brown-haired and one gray-haired (litter 1), and the other with three brown-haired and two gray-haired (litter 2). We select a litter at random and then select an offspring at random from the selected litter.

1. What is the probability that the animal chosen is gray-haired?

Let´s denote:

|s|= 6

Under the assumption of independence \(\bf{P(G\cap Lit_{1}) = P(G) * P(Lit_{1})}\)

\[ \bf{P(G)}= P(G \cap Lit1) + P(G \cap Lit 2) = \frac{1}{6} + \frac{1}{5} = \frac{11}{30} \] b. Given that a gray-haired offspring was selected, what is the probability that the sampling was from litter 2? \[ \bf{P(Lit2| G)}= \frac {P(Lit2 \cap G)} {P(G)} = \frac{P(Lit) *P(G)}{P(G)} = \frac{1}{2} \]