unit 11 homework 3 conditional probability

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8.4.1: Conditional Probability (Exercises)

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• Rupinder Sekhon and Roberta Bloom
• De Anza College

SECTION 8.4 PROBLEM SET: CONDITIONAL PROBABILITY

Questions 1 - 4: Do these problems using the conditional probability formula: \(P(A | B)=\frac{P(A \cap B)}{P(B)}\).

Questions 5 - 8 refer to the following: The table shows the distribution of Democratic and Republican U.S. Senators by gender in the 114 th Congress as of January 2015.

Use this table to determine the following probabilities:

Do the following conditional probability problems.

At a college, 72% of courses have final exams and 46% of courses require research papers. 32% of courses have both a research paper and a final exam. Let \(F\) be the event that a course has a final exam and \(R\) be the event that a course requires a research paper.

Consider a family of three children. Find the following probabilities.

Questions 21 - 26 refer to the following: The table shows highest attained educational status for a sample of US residents age 25 or over:

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11.4: Conditional Probability

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• Page ID 34505

• Rupinder Sekhon and Roberta Bloom
• De Anza College

Learning Objectives

In this section, you will learn to:

• recognize situations involving conditional probability
• calculate conditional probabilities

Suppose a friend asks you the probability that it will snow today.

If you are in Boston, Massachusetts in the winter, the probability of snow today might be quite substantial. If you are in Cupertino, California in summer, the probability of snow today is very tiny, this probability is pretty much 0.

• \(\mathrm{A}\) = the event that it will snow today
• \(\mathrm{B}\) = the event that today you are in Boston in wintertime
• \(\mathrm{C}\) = the event that today you are in Cupertino in summertime

Because the probability of snow is affected by the location and time of year, we can’t just write \(\mathrm{P(A)}\) for the probability of snow. We need to indicate the other information we know -location and time of year. We need to use conditional probability .

The event we are interested in is event \(\mathrm{A}\) for snow. The other event is called the condition, representing location and time of year in this case.

We represent conditional probability using a vertical line | that means “if”, or “given that”, or “if we know that”. The event of interest appears on the left of the |. The condition appears on the right side of the |.

The probability it will snow given that (if) you are in Boston in the winter is represented by \(\mathbf{P} \left( \mathbf{A} | \mathbf{B}\right) \). In this case, the condition is \(\mathrm{B}\).

The probability that it will snow given that (if) you are in Cupertino in the summer is represented by \(\mathbf{P}(\mathbf{A} | \mathbf{C})\). In this case, the condition is \(\mathrm{C}\).

Now, let’s examine a situation where we can calculate some probabilities.

Suppose you and a friend play a game that involves choosing a single card from a well-shuffled deck. Your friend deals you one card, face down, from the deck and offers you the following deal: If the card is a king, he will pay you $5, otherwise, you pay him $1. Should you play the game?

You reason in the following manner. Since there are four kings in the deck, the probability of obtaining a king is 4/52 or 1/13. So, probability of not obtaining a king is 12/13. This implies that the ratio of your winning to losing is 1 to 12, while the payoff ratio is only $1 to $5. Therefore, you determine that you should not play.

But consider the following scenario. While your friend was dealing the card, you happened to get a glance of it and noticed that the card was a face card. Should you, now, play the game?

Since there are 12 face cards in the deck, the total elements in the sample space are no longer 52, but just 12. This means the chance of obtaining a king is 4/12 or 1/3. So your chance of winning is 1/3 and of losing 2/3. This makes your winning to losing ratio 1 to 2 which fares much better with the payoff ratio of $1 to $5. This time, you determine that you should play.

In the second part of the above example, we were finding the probability of obtaining a king knowing that a face card had shown. This is an example of conditional probability . Whenever we are finding the probability of an event \(\mathrm{E}\) under the condition that another event \(\mathrm{F}\) has happened, we are finding conditional probability.

The symbol \(\mathrm{P(E | F)}\) denotes the problem of finding the probability of \(\mathrm{E}\) given that \(\mathrm{F}\) has occurred. We read \(\mathrm{P(E | F)}\) as "the probability of \(\mathrm{E}\), given \(\mathrm{F}\)."

Example \(\PageIndex{1}\)

A family has three children. Find the conditional probability of having two boys and a girl given that the first born is a boy.

Let event \(\mathrm{E}\) be that the family has two boys and a girl, and \(\mathrm{F}\) that the first born is a boy.

First, we the sample space for a family of three children as follows.

\[S = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\} \nonumber \]

Since we know the first born is a boy, our possibilities narrow down to four outcomes: BBB, BBG, BGB, and BGG.

Among the four, BBG and BGB represent two boys and a girl.

Therefore, \(\mathrm{P(E | F)}\) = 2/4 or 1/2.

Example \(\PageIndex{2}\)

One six sided die is rolled once.

• Find the probability that the result is even.
• Find the probability that the result is even given that the result is greater than three.

The sample space is \(\mathrm{S} = {1,2,3,4,5,6}\)

Let event \(\mathrm{E}\) be that the result is even and \(\mathrm{T}\) be that the result is greater than 3.

a. \(\mathrm{P(E)}\) = 3/6 because \(\mathrm{E} = {2,4,6}\)

b. Because \(T = {4,5,6}\), we know that 1, 2, 3 cannot occur; only outcomes 4, 5, 6 are possible. Therefore of the values in \(\mathrm{E}\), only 4, 6 are possible.

Therefore, \(\mathrm{P(E|T)}\) = 2/3

Example \(\PageIndex{3}\)

A fair coin is tossed twice.

• Find the probability that the result is is two heads.
• Find the probability that the result is two heads given that at least one head is obtained.

The sample space is \(S = {HH, HT, TH, TT}\)

Let event \(\mathrm{E}\) be that the two heads are obtained and \(\mathrm{F}\) be at least one head is obtained

a. \(\mathrm{P(E)}\) = 1/4 because \(\mathrm{E} = {HH}\) and the sample space \(\mathrm{S}\) has 4 outcomes.

b. \(\mathrm{F} = {HH, HT, TH}\). Since at least one head was obtained, TT did not occur. We are interested in the probability event \(\mathrm{E}={HH}\) out of the 3 outcomes in the reduced sample space F.

Therefore, \(\mathrm{P(E|F)}\) = 1/3

Let us now develop a formula for the conditional probability \(\mathrm{P(E | F)}\).

Suppose an experiment consists of \(n\) equally likely events. Further suppose that there are \(m\) elements in \(\mathrm{F}\), and \(c\) elements in \(\mathrm{E} \cap \mathrm{F}\), as shown in the following Venn diagram.

If the event \(\mathrm{F}\) has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset \(\mathrm{F}\). Therefore, we only look at the set \(\mathrm{F}\) and at nothing outside of \(\mathrm{F}\). Since \(\mathrm{F}\) has \(m\) elements, the denominator in the calculation of \(\mathrm{P(E | F)}\) is \(m\). We may think that the numerator for our conditional probability is the number of elements in \(\mathrm{E}\). But clearly we cannot consider the elements of \(\mathrm{E}\) that are not in \(\mathrm{F}\). We can only count the elements of \(\mathrm{E}\) that are in \(\mathrm{F}\), that is, the elements in \(\mathrm{E} \cap \mathrm{F}\). Therefore,

\[\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{c}}{\mathrm{m}} \nonumber \]

Dividing both the numerator and the denominator by \(n\), we get

\[\mathrm{P(E | F)}=\frac{c / n}{m / n} \nonumber \]

But \(c/n = \mathrm{P}(\mathrm{E} \cap \mathrm{F})\), and \(m/n = \mathrm{P}(\mathrm{F})\).

Substituting, we derive the following formula for \(\mathrm{P}(\mathrm{E} | \mathrm{F})\).

Conditional Probability Rule

For two events \(\mathrm{E}\) and \(\mathrm{F}\), the probability of "\(\mathrm{E}\) Given \(\mathrm{F}\)" is

\[\mathbf{P}(\mathbf{E} | \mathbf{F})=\frac{\mathbf{P}(\mathbf{E} \cap \mathbf{F})}{\mathbf{P}(\mathbf{F})} \nonumber \]

Example \(\PageIndex{4}\)

A single die is rolled. Use the above formula to find the conditional probability of obtaining an even number given that a number greater than three has shown.

Let \(\mathrm{E}\) be the event that an even number shows, and \(\mathrm{F}\) be the event that a number greater than three shows. We want \(\mathrm{P}(\mathrm{E} | \mathrm{F})\).

\(\mathrm{E} = {2, 4, 6}\) and \(\mathrm{F} = {4, 5, 6}\). Which implies, \(\mathrm{E} \cap \mathrm{F} = { 4, 6}\)

Therefore, \(\mathrm{P}(\mathrm{F})\) = 3/6, and \(\mathrm{P}(\mathrm{E} \cap \mathrm{F})\) = 2/6

\[P(E | F)=\frac{P(E \cap F)}{P(F)}=\frac{2 / 6}{3 / 6}=\frac{2}{3} \nonumber. \nonumber \]

Example \(\PageIndex{5}\)

The following table shows the distribution by gender of students at a community college who take public transportation and the ones who drive to school.

The events \(\mathrm{M}, \mathrm{F}, \mathrm{T}\), and \(\mathrm{D}\) are self explanatory. Find the following probabilities.

• \(\mathrm{P}(\mathrm{D} | \mathrm{M})\)
• \(\mathrm{P}(\mathrm{F} | \mathrm{D})\)
• \(\mathrm{P}(\mathrm{M} | \mathrm{T})\)

Conditional probabilities can often be found directly from a contingency table. If the condition corresponds to only one row or only one column in the table, then you can ignore the rest of the table and read the conditional probability directly from the row or column indicated by the condition.

• The condition is event \(\mathrm{M}\); we can look at only the “Male” column of the table and ignore the rest of the table: \(\mathrm{P}(\mathrm{D} | \mathrm{M})=\frac{39}{47}\).
• The condition is event \(\mathrm{D}\); we can look at only the “Drive” row of the table and ignore the rest of the table: \(\mathrm{P}(\mathrm{F} | \mathrm{D})=\frac{40}{79}\).
• The condition is event \(\mathrm{T}\); we can look at only the “Public Transportation” row of the table and ignore the rest of the table: \(\mathrm{P}(\mathrm{M} | \mathrm{T})=\frac{8}{21}\).

We use the conditional probability formula \(\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P(F)}}\).

• \[P(D | M)=\frac{P(D \cap M)}{P(M)}=\frac{39 / 100}{47 / 100}=\frac{39}{47} \nonumber . \nonumber \]
• \[P(F | D)=\frac{P(F \cap D)}{P(D)}=\frac{40 / 100}{79 / 100}=\frac{40}{79} \nonumber . \nonumber \]
• \[\mathrm{P}(\mathrm{M} | \mathrm{T})=\frac{P(M \cap T)}{P(T)}=\frac{8 / 100}{21 / 100}=\frac{8}{21} \nonumber \]

Example \(\PageIndex{6}\)

Given \(\mathrm{P(E)}\) = .5, \(\mathrm{P}(\mathrm{F})\) = .7, and \(\mathrm{P}(\mathrm{E} \cap \mathrm{F})\) = .3. Find the following:

• \(\mathrm{P}(\mathrm{E} | \mathrm{F})\)
• \(\mathrm{P}(\mathrm{F} | \mathrm{E})\)

We use the conditional probability formula.

• \(P(E | F)=\frac{P(E \cap F)}{P(F)}=\frac{3}{7}=\frac{3}{7}\)
• \(P(F | E)=\frac{P(E \cap F)}{P(E)}=.3 / .5=3 / 5 \)

Example \(\PageIndex{7}\)

\(\mathrm{E}\) and \(\mathrm{F}\) are mutually exclusive events such that \(\mathrm{P(E)}\) = .4, \(\mathrm{P(F)}\) = .9. Find \(\mathrm{P(E | F)}\).

\(\mathrm{E}\) and \(\mathrm{F}\) are mutually exclusive, so \(\mathrm{P}(\mathrm{E} \cap \mathrm{F})\) = 0. Therefore \(\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{0}{9}=0\).

Example \(\PageIndex{8}\)

Given \(\mathrm{P}(\mathrm{F} | \mathrm{E})\) = .5, and \(\mathrm{P}(\mathrm{E} \cap \mathrm{F})\) = .3. Find \(\mathrm{P}(\mathrm{E})\).

Using the conditional probability formula \(\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}\), we get

\[\mathrm{P}(\mathrm{F} | \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})} \nonumber \]

Substituting and solving:

\[.5=\frac{.3}{\mathrm{P}(\mathrm{E})} \quad \text { or } \quad \mathrm{P}(\mathrm{E})=3 / 5 \nonumber \]

Example \(\PageIndex{9}\)

In a family of three children, find the conditional probability of having two boys and a girl, given that the family has at least two boys.

Let event \(\mathrm{E}\) be that the family has two boys and a girl, and let \(\mathrm{F}\) be the probability that the family has at least two boys. We want \(\mathrm{P}(\mathrm{E} | \mathrm{F})\).

We list the sample space along with the events \(\mathrm{E}\) and \(\mathrm{F}\).

\begin{aligned} &\mathrm{S}=\{\mathrm{BBB}, \mathrm{BBG}, \mathrm{BGB}, \mathrm{BGG}, \mathrm{GBB}, \mathrm{GBG}, \mathrm{GGB}, \mathrm{GGG}\}\\ &\mathrm{E}=\{\mathrm{BBG}, \mathrm{BGB}, \mathrm{GBB}\} \text { and } \mathrm{F}=\{\mathrm{BBB}, \mathrm{BBG}, \mathrm{BGB}, \mathrm{GBB}\}\\ &\mathrm{E} \cap \mathrm{F}=\{\mathrm{BBG}, \mathrm{BGB}, \mathrm{GBB}\} \end{aligned}

Therefore, \(\mathrm{P}(\mathrm{F})\) = 4/8, and \(\mathrm{P}(\mathrm{E} \cap \mathrm{F})\) = 3/8, and

\[\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})} =\frac{3/8}{4/8} =\frac{3}{4} \nonumber. \nonumber \]

Example \(\PageIndex{10}\)

At a community college 65% of the students subscribe to Amazon Prime, 50% subscribe to Netflix, and 20% subscribe to both. If a student is chosen at random, find the following probabilities:

• the student subscribes to Amazon Prime given that he subscribes to Netflix
• the student subscribes to Netflix given that he subscribes to Amazon Prime

Let \(\mathrm{A}\) be the event that the student subscribes to Amazon Prime, and \(\mathrm{N}\) be the event that the student subscribes to Netflix.

First identify the probabilities and events given in the problem.

\(\mathrm{P}\)(student subscribes to Amazon Prime) = \(\mathrm{P(A)}\) = 0.65

\(\mathrm{P}\)(student subscribes to Netflix) = \(\mathrm{P(N)}\) = 0.50

\(\mathrm{P}\)(student subscribes to both Amazon Prime and Netflix) = \(\mathrm{P(A \cap N)}\) = 0.20

Then use the conditional probability rule:

• \(\mathrm{P}(\mathrm{A} | \mathrm{N})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{N})}{\mathrm{P}(\mathrm{N})} = \frac{.20}{.50} = \frac{2}{5}\)
• \(\mathrm{P}(\mathrm{N} | \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{N})}{\mathrm{P}(\mathrm{A})} = \frac{.20}{.65} = \frac{4}{13}\)

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Unit 11 Homework 3 Conditional Probability

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3.2: Problems on Conditional Probability

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• Page ID 10875

• Paul Pfeiffer
• Rice University

Exercise \(\PageIndex{1}\)

Given the following data:

\(P(A) = 0.55\), \(P(AB) = 0.30\), \(P(BC) = 0.20\), \(P(A^c \cup BC) = 0.55\), \(P(A^c BC^c) = 0.15\)

Determine, if possible, the conditional probability \(P(A^c|B) = P(A^cB)/P(B)\).

Exercise \(\PageIndex{2}\)

In Exercise 11 from "Problems on Minterm Analysis," we have the following data: A survey of a represenative group of students yields the following information:

• 52 percent are male
• 85 percent live on campus
• 78 percent are male or are active in intramural sports (or both)
• 30 percent live on campus but are not active in sports
• 32 percent are male, live on campus, and are active in sports
• 8 percent are male and live off campus
• 17 percent are male students inactive in sports

Let A = male, B = on campus, C = active in sports.

• A student is selected at random. He is male and lives on campus. What is the (conditional) probability that he is active in sports?
• A student selected is active in sports. What is the(conditional) probability that she is a female who lives on campus?

Exercise \(\PageIndex{3}\)

In a certain population, the probability a woman lives to at least seventy years is 0.70 and is 0.55 that she will live to at least eighty years. If a woman is seventy years old, what is the conditional probability she will survive to eighty years? Note that if \(A \subset B\) then \(P(AB) = P(A)\).

Let \(A=\) event she lives to seventy and \(B=\) event she lives to eighty. Since \(B \subset A\), \(P(B|A) = P(AB)/P(A) = P(B)/P(A) = 55/70\).

Exercise \(\PageIndex{4}\)

From 100 cards numbered 00, 01, 02, \(\cdot\cdot\cdot\), 99, one card is drawn. Suppose A i is the event the sum of the two digits on a card is \(i\), \(0 \le i \le 18\), and \(B_j\) is the event the product of the two digits is \(j\). Determine \(P(A_i|B_0)\) for each possible \(i\).

\(B_0\) is the event one of the first ten is draw. \(A_i B_0\) is the event that the card with numbers \(0i\) is drawn. \(P(a_i|B_0) = (1/100)/(1/10) = 1/10\) for each \(i\), 0 through 9.

Exercise \(\PageIndex{5}\)

Two fair dice are rolled.

• What is the (conditional) probability that one turns up two spots, given they show different numbers?
• What is the (conditional) probability that the first turns up six, given that the sum is \(k\), for each \(k\) from two through 12?
• What is the (conditional) probability that at least one turns up six, given that the sum is \(k\), for each \(k\) from two through 12?

a. There are \(6 \times 5\) ways to choose all different. There are \(2 \times 5\) ways that they are different and one turns up two spots. The conditional probability is 2/6.

b. Let \(A_6\) = event first is a six and \(S_k = \) event the sum is \(k\). Now \(A_6S_k = \emptyset\) for \(k \le 6\). A table of sums shows \(P(A_6S_k) = 1/36\) and \(P(S_k) = 6/36, 5/36, 4/36, 3/36, 2/36, 1/36\) for \(k = 7\) through 12, respectively. Hence \(P(A_6|S_k) = 1/6, 1/5. 1/4, 1/3. 1/2, 1\), respectively.

c. If \(AB_6\) is the event at least one is a six, then \(AB_6S_k) = 2/36\) for \(k = 7\) through 11 and \(P(AB_6S_12) = 1/36\). Thus, the conditional probabilities are 2/6, 2/5, 2/4, 2/3, 1, 1, respectively.

Exercise \(\PageIndex{6}\)

Four persons are to be selected from a group of 12 people, 7 of whom are women.

• What is the probability that the first and third selected are women?
• What is the probability that three of those selected are women?
• What is the (conditional) probability that the first and third selected are women, given that three of those selected are women?

\(P(W_1W_3) = P(W_1W_2W_3) + P(W_1W_2^c W_3) = \dfrac{7}{12} \cdot \dfrac{6}{11} \cdot \dfrac{5}{10} + \dfrac{7}{12} \cdot \dfrac{5}{11} \cdot \dfrac{6}{10} = \dfrac{7}{22}\)

Exercise \(\PageIndex{7}\)

Twenty percent of the paintings in a gallery are not originals. A collector buys a painting. He has probability 0.10 of buying a fake for an original but never rejects an original as a fake, What is the (conditional) probability the painting he purchases is an original?

Let \(B=\) the event the collector buys, and \(G=\) the event the painting is original. Assume \(P(B|G) = 1\) and \(P(B|G^c) = 0.1\). If \(P(G) = 0.8\), then

\(P(G|B) = \dfrac{P(GB)}{P(B)} = \dfrac{P(B|G) P(G)}{P(B|G)P(G) + P(B|G^c)P(G^c)} = \dfrac{0.8}{0.8 + 0.1 \cdot 0.2} = \dfrac{40}{41}\)

Exercise \(\PageIndex{8}\)

Five percent of the units of a certain type of equipment brought in for service have a common defect. Experience shows that 93 percent of the units with this defect exhibit a certain behavioral characteristic, while only two percent of the units which do not have this defect exhibit that characteristic. A unit is examined and found to have the characteristic symptom. What is the conditional probability that the unit has the defect, given this behavior?

Let \(D=\) the event the unit is defective and \(C=\) the event it has the characteristic. Then \(P(D) = 0.05\), \(P(C|D) = 0.93\), and \(P(C|D^c) = 0.02\).

\(P(D|C) = \dfrac{P(C|D) P(D)}{P(C|D) P(D) + P(C|D^c) P(D^c)} = \dfrac{0.93 \cdot 0.05}{0.93 \cdot 0.05 + 0.02 \cdot 0.95} = \dfrac{93}{131}\)

Exercise \(\PageIndex{9}\)

A shipment of 1000 electronic units is received. There is an equally likely probability that there are 0, 1, 2, or 3 defective units in the lot. If one is selected at random and found to be good, what is the probability of no defective units in the lot?

Let \(D_k =\) the event of \(k\) defective and \(G\) be the event a good one is chosen.

\(P(D_0|G) = \dfrac{P(G|D_0) P(D_0)}{P(G|D_0) P(D_0) + P(G|D_1) P(D_1) + P(G|D_2) P(D_2) + P(G|D_3) P(D_3)}\)

\(= \dfrac{1 \cdot 1/4}{(1/4)(1 + 999/1000 + 998/1000 + 997/1000)} = \dfrac{1000}{3994}\)

Exercise \(\PageIndex{10}\)

\(P(S_i|E_j)\)

• Suppose a person has a university education (no graduate study). What is the (conditional) probability that he or she will make $25,000 or more?
• Find the total probability that a person's income category is at least as high as his or her educational level.

a. \(P(E_3S_3) = P(S_3|E_3)P(E_3) = 0.45 \cdot 0.05 = 0.0225\)

b. \(P(S_2 \vee S_3|E_2) = 0.80 + 0.10 = 0.90\)

c. \(p = (0.85 + 0.10 + 0.05) \cdot 0.65 + (0.80 + 0.10) \cdot 0.30 + 0.45 \cdot 0.05 = 0.9425\)

Exercise \(\PageIndex{11}\)

In a survey, 85 percent of the employees say they favor a certain company policy. Previous experience indicates that 20 percent of those who do not favor the policy say that they do, out of fear of reprisal. What is the probability that an employee picked at random really does favor the company policy? It is reasonable to assume that all who favor say so.

\(P(S) = 0.85\), \(P(S|F^c) = 0.20\). Also, reasonable to assume \(P(S|F) = 1\).

\(P(S) = P(S|F) P(F) + P(S|F^c) [1 - P(F)]\) implies \(P(F) = \dfrac{P(S) - P(S|F^c)}{1 - P(S|F^c)} = \dfrac{13}{16}\)

Exercise \(\PageIndex{12}\)

\(\dfrac{D^c|T}{P(D|T)} = \dfrac{P(T|D^c)P(D^c)}{P(T|D)P(D)} = \dfrac{0.98 \cdot 0.99}{0.05 \cdot 0.01} = \dfrac{9702}{5}\)

\(P(D^c|T) = \dfrac{9702}{9707} = 1 - \dfrac{5}{9707}\)

Exercise \(\PageIndex{13}\)

Five boxes of random access memory chips have 100 units per box. They have respectively one, two, three, four, and five defective units. A box is selected at random, on an equally likely basis, and a unit is selected at random therefrom. It is defective. What are the (conditional) probabilities the unit was selected from each of the boxes?

\(H_i =\) the event from box \(i\). \(P(H_i) = 1/5\) and \(P(D|H_i) = i/100\).

\(P(H_i|D) = \dfrac{P(D|H_i) P(H_i)}{\sum P(D|H_i) P(H_j)} = i/15\), \(1 \le i \le 5\)

Exercise \(\PageIndex{14}\)

Two percent of the units received at a warehouse are defective. A nondestructive test procedure gives two percent false positive indications and five percent false negative. Units which fail to pass the inspection are sold to a salvage firm. This firm applies a corrective procedure which does not affect any good unit and which corrects 90 percent of the defective units. A customer buys a unit from the salvage firm. It is good. What is the (conditional) probability the unit was originally defective?

Let \(T\) = event test indicates defective, \(D\) = event initially defective, and \(G =\) event unit purchased is good. Data are

\(P(D) = 0.02\), \(P(T^c|D) = 0.02\), \(P(T|D^c) = 0.05\), \(P(GT^c) = 0\),

\(P(G|DT) = 0.90\), \(P(G|D^cT) = 1\)

\(P(D|G) = \dfrac{P(GD)}{P(G)}\), \(P(GD) = P(GTD) = P(D) P(T|D) P(G|TD)\)

\(P(G) = P(GT) = P(GDT) + P(GD^c T) = P(D) P(T|D) P(G|TD) + P(D^c) P(T|D^c) P(G|TD^c)\)

\(P(D|G) = \dfrac{0.02 \cdot 0.98 \cdot 0.90}{0.02 \cdot 0.98 \cdot 0.90 + 0.98 \cdot 0.05 \cdot 1.00} = \dfrac{441}{1666}\)

Exercise \(\PageIndex{15}\)

At a certain stage in a trial, the judge feels the odds are two to one the defendent is guilty. It is determined that the defendent is left handed. An investigator convinces the judge this is six times more likely if the defendent is guilty than if he were not. What is the likelihood, given this evidence, that the defendent is guilty?

Let \(G\) = event the defendent is guilty, \(L\) = the event the defendent is left handed. Prior odds: \(P(G)/P(G^c) = 2\). Result of testimony: \(P(L|G)/P(L|G^c) = 6\).

\(\dfrac{P(G|L)}{P(G^c|L)} = \dfrac{P(G)}{P(G^c)} \cdot \dfrac{P(L|G)}{P(L|G^c)} = 2 \cdot 6 = 12\)

\(P(G|L) = 12/13\)

Exercise \(\PageIndex{16}\)

Show that if \(P(A|C) > P(B|C)\) and \(P(A|C^c) > P(B|C^c)\), then \(P(A) > P(B)\). Is the converse true? Prove or give a counterexample.

\(P(A) = P(A|C) P(C) + P(A|C^c) P(C^c) > P(B|C) P(C) + P(B|C^c) P(C^c) = P(B)\).

The converse is not true. Consider \(P(C) = P(C^c) = 0.5\), \(P(A|C) = 1/4\).

\(P(A|C^c) = 3/4\), \(P(B|C) = 1/2\), and \(P(B|C^c) = 1/4\). Then

\(1/2 = P(A) = \dfrac{1}{2} (1/4 + 3/4) > \dfrac{1}{2} (1/2 + 1/4) = P(B) = 3/8\)

But \(P(A|C) < P(B|C)\).

Exercise \(\PageIndex{17}\)

Since \(P(\cdot |B)\) is a probability measure for a given \(B\), we must have \(P(A|B) + P(A^c|B) = 1\). Construct an example to show that in general \(P(A|B) + P(A|B^c) \ne 1\).

Suppose \(A \subset B\) with \(P(A) < P(B)\). Then \(P(A|B) = P(A)/P(B) < 1\) and \(P(A|B^c) = 0\) so the sum is less than one.

Exercise \(\PageIndex{18}\)

Use property ( CP4 ) to show

a. \(P(A|B) > P(A)\) iff \(P(A|B^c) < P(A)\)

b. \(P(A^c|B) > P(A^c)\) iff \(P(A|B) < P(A)\)

c. \(P(A|B) > P(A)\) iff \(P(A^c|B^c) > P(A^c)\)

a. \(P(A|B) > P(A)\) iff \(P(AB) > P(A) P(B)\) iff \(P(AB^c) < P(A) P(B^c)\) iff \(P(A|B^c) < P(A)\)

b. \(P(A^c|B) > P(A^c)\) iff \(P(A^c B) > P(A^c) P(B)\) iff \(P(AB) < P(A) P(B)\) iff \(P(A|B) < P(A)\)

c. \(P(A|B) > P(A)\) iff \(P(AB) > P(A) P(B)\) iff \(P(A^c B^c) > P(A^c) P(B^c)\) iff \(P(A^c|B^c) > P(A^c)\)

Exercise \(\PageIndex{19}\)

Show that \(P(A|B) \ge (P(A) + P(B) - 1)/P(B)\).

\(1 \ge P(A \cup B) = P(A) + P(B) - P(AB) = P(A) + P(B) - P(A|B) P(B)\). Simple algebra gives the desired result.

Exercise \(\PageIndex{20}\)

Show that \(P(A|B) = P(A|BC) P(C|B) + P(A|BC^c) P(C^c|B)\).

\(P(A|B) = \dfrac{P(AB)}{P(B)} = \dfrac{P(ABC) + P(ABC^c)}{P(B)}\)

\(= \dfrac{P(A|BC) P(BC) + P(A|BC^c) P(BC^c)}{P(B)} = P(A|BC) P(C|B) + P(A|BC^c) P(C^c|B)\)

Exercise \(\PageIndex{21}\)

\(P(A|B) = 1\), \(P(A|B^c) = 1/n\), \(P(B) = p\)

\(P(B|A) = \dfrac{P(A|B) P(B)}{P(A|B) P(B) +P(A|B^c) P(B^c)} = \dfrac{p}{p + \dfrac{1}{n} (1 - p)} = \dfrac{np}{(n - 1) p + 1}\)

\(\dfrac{P(B|A)}{P(B)} = \dfrac{n}{np + 1 - p}\) increases from 1 to \(1/p\) as \(n \to \infty\)

Exercise \(\PageIndex{22}\)

Polya's urn scheme for a contagious disease . An urn contains initially \(b\) black balls and \(r\) red balls \((r + b = n)\). A ball is drawn on an equally likely basis from among those in the urn, then replaced along with \(c\) additional balls of the same color. The process is repeated. There are \(n\) balls on the first choice, \(n + c\) balls on the second choice, etc. Let \(B_k\) be the event of a black ball on the \(k\)th draw and \(R_k\) be the event of a red ball on the \(k\)th draw. Determine

a. \(P(B_2|R_1)\) b. \(P(B_1B_2)\) c. \(P(R_2)\) d. \(P(B_1|R_2)\)

a. \(P(B_2|R_1) = \dfrac{b}{n + c}\)

b. \(P(B_1B_2) = P(B_2) P(B_2|B_1) = \dfrac{b}{n} \cdot \dfrac{b + c}{n + c}\)

c. \(P(R_2) P(R_2|R_1) P(R_1) + P(R_2|B_1) P(B_1)\)

\(= \dfrac{r + c}{n + c} \cdot \dfrac{r}{n} + \dfrac{r}{n + c} \cdot \dfrac{b}{n} = \dfrac{r(r + c + b)}{n(n + c)}\)

d. \(P(B_1|R_2) = \dfrac{P(R_2|B_1) P(B_1)}{P(R_2)}\) with \(P(R_2|B_1) P(B_1) = \dfrac{r}{n + c} \cdot \dfrac{b}{n}\). Using (c), we have

\(P(B_1|R_2) = \dfrac{b}{r + b + c} = \dfrac{b}{n + c}\)

Algebra 2 Common Core

By hall, prentice, chapter 11 - probability and statistics - 11-1 permutations and combinations - lesson check - page 678: 3, work step by step, update this answer.

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Unit 8: Probability & Statistics Name: _ Date:__Per:_Homework 3: Theoretical vs. Experimental Probability ** This is a 2-page document! ** Give each probability as a simplified fraction, decimal, and percent 1. A number between 1 and 3 is a) Find and compare the theoretical probability and chosen 30 times. Results are shown experimental probability of choosing a 2. in the table below. Theoretical: Experimental: Compare: 2. The spinner below is spun 60 times. a) Find and compare the theoretical probability and Results are shown in the table experimental probability of spinning an A

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