Quantitative Reasoning Chapter 3 MTH105

 In chapter 3 we cover probability which is a fundamental concept in quantitative reasoning, and it deals with the likelihood of events occurring. Here are some key principles and methods related to probability in quantitative reasoning:

Probability Basics:

Probability is expressed as a number between 0 and 1, where 0 represents an event that is impossible, and 1 represents an event that is certain.

The probability of an event A is denoted as P(A).

Probability of an Event:

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

P(A) = Number of Favorable Outcomes / Total Number of Possible Outcomes

Probability Distributions:

Probability distributions describe the likelihood of different outcomes in a random experiment or data set. Common probability distributions include the uniform distribution, binomial distribution, and normal distribution.

Conditional Probability:

Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A | B) and is calculated as:

P(A | B) = P(A and B) / P(B)

Bayes' Theorem:

Bayes' Theorem is used to update probabilities based on new information. It's particularly useful in situations where you have prior probabilities and want to update them with new evidence.

Combinatorics:

Combinatorial methods, such as permutations and combinations, are often used to calculate probabilities in situations involving arrangements and selections of items.

Expected Value:

The expected value (or mean) of a random variable is the weighted average of all possible values of the variable, where each value is weighted by its probability of occurring.

E(X) = Σ [x * P(x)]

Probability in Data Analysis:

In data analysis, probability is used to make predictions, conduct hypothesis testing, and perform statistical inference.

Simulation:

Probability can also be explored by simulation, where random processes are simulated using computer programs to estimate probabilities and outcomes.

Unit 3 Lesson:

A group of people were asked if they had run a red light in the last year. 331 responded "yes", and 129 responded "no".

Find the probability that if a person is chosen at random, they have run a red light in the last year.

$0.720$0.720​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Give your answer as a fraction or decimal accurate to at least 3 decimal places

To find the probability that a randomly chosen person has run a red light in the last year, you can use the following formula:

Probability (P) = Number of favorable outcomes / Total number of possible outcomes

In this case, the favorable outcomes are the people who responded "yes," which is 331, and the total number of possible outcomes is the sum of the people who responded "yes" and "no," which is 331 (yes) + 129 (no) = 460.

So, the probability is:

P = 331 (favorable outcomes) / 460 (total possible outcomes)

P ≈ 0.7196

Then we round it 3 decimal places  0.720

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	19	9	6	34
Female	2	14	18	34
Total	21	23	24	68

If one student was chosen at random,

find the probability that the student was female.

$0.5$0.5​Correct  

Give your answer as a fraction or decimal (4 places after the decimal if needed).

To find the probability that a randomly chosen student is female, you need to use the information from the table that summarizes the grades and gender of the students.

The total number of students is given as 68, and you can see from the table that there are 34 female students.

So, the probability that a randomly chosen student is female is:

Probability (Female) = Number of Female Students / Total Number of Students

Probability (Female) = 34 / 68

Probability (Female) = 1/2

So, the probability that the student chosen at random is female is 1/2 or 0.5000 rounded to 0.5

A die is rolled. Find the probability of the given event.
(a) The number showing is a 3;
The probability is : $0.1667$0.1667​Correct  
(b) The number showing is an even number;
The probability is : $0.5$0.5​Correct  
(c) The number showing is greater than 3;
The probability is : $0.50$0.50​Correct  

Round answers 4 places after the decimal if needed.

Let's find the probabilities for each of the events:

(a) The number showing is a 3:

There is only one favorable outcome (rolling a 3).

There are six possible outcomes (rolling any number from 1 to 6).

Probability (rolling a 3) = 1 (favorable outcome) / 6 (total possible outcomes)

Probability (rolling a 3) = 1/6

(b) The number showing is an even number:

Even numbers on a standard six-sided die are 2, 4, and 6.

There are three favorable outcomes (rolling a 2, 4, or 6).

There are six possible outcomes (rolling any number from 1 to 6).

Probability (rolling an even number) = 3 (favorable outcomes) / 6 (total possible outcomes)

Probability (rolling an even number) = 3/6

(c) The number showing is greater than 3:

Numbers greater than 3 on a standard six-sided die are 4, 5, and 6.

There are three favorable outcomes (rolling a 4, 5, or 6).

There are six possible outcomes (rolling any number from 1 to 6).

Probability (rolling a number greater than 3) = 3 (favorable outcomes) / 6 (total possible outcomes)

Probability (rolling a number greater than 3) = 3/6

Now, let's simplify these fractions:

(a) Probability (rolling a 3) = 1/6

(b) Probability (rolling an even number) = 3/6 = 1/2

(c) Probability (rolling a number greater than 3) = 3/6 = 1/2

So, rounded to four decimal places:

We divide the fractions to get the decimals: 

(a) Probability (rolling a 3) 1/6= 0.1667

(b) Probability (rolling an even number) 1/2= 0.5000

(c) Probability (rolling a number greater than 3) 1/2= 0.5000

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	7	19	5	31
Female	17	8	20	45
Total	24	27	25	76

If one student is chosen at random,

Find the probability that the student did NOT get an "C"

Round answers 4 places after the decimal if needed.

$0.6711$0.6711​Correct  

To find the probability that a randomly chosen student did NOT get a "C," you can use the information from the table that summarizes the grades and gender of the students.

The total number of students who did NOT get a "C" can be calculated by subtracting the number of students who received a "C" from the total number of students. In this case:

Total number of students who did NOT get a "C" = Total number of students - Number of students who got a "C"

Total number of students = 76

Number of students who got a "C" = 25

Total number of students who did NOT get a "C" = 76 - 25 = 51

Now, you can find the probability that a randomly chosen student did NOT get a "C" using the following formula:

Probability (not getting a "C") = Number of students who did NOT get a "C" / Total number of students

Probability (not getting a "C") = 51 / 76

Now, calculate the probability, rounding to four decimal places:

Probability (not getting a "C") ≈ 0.6711

So, the probability that a randomly chosen student did NOT get a "C" is approximately 0.6711 when rounded to four decimal places.

A die is rolled twice. What is the probability of showing a 5 on the first roll and an even number on the second roll?

Your answer is : $\frac{1}{12}$112​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Express answer as a fraction.

To find the probability of rolling a 5 on the first roll and an even number on the second roll of a fair six-sided die, you can break it down into two separate probabilities:

The probability of rolling a 5 on the first roll:

There is one favorable outcome (rolling a 5) out of a total of six possible outcomes (rolling any number from 1 to 6).

The probability of rolling an even number on the second roll:

There are three favorable outcomes (rolling a 2, 4, or 6) out of a total of six possible outcomes.

Now, you can find the overall probability by multiplying the probabilities of the two events:

Probability (rolling a 5 on the first roll and an even number on the second roll) = Probability (rolling a 5 on the first roll) * Probability (rolling an even number on the second roll)

Probability (rolling a 5 on the first roll and an even number on the second roll) = (1/6) * (3/6)

Now, calculate the product:

Probability (rolling a 5 on the first roll and an even number on the second roll) = (1/6) * (3/6) = 3/36

You can simplify the fraction:

Probability (rolling a 5 on the first roll and an even number on the second roll) = 1/12

So, the probability of rolling a 5 on the first roll and an even number on the second roll of a fair six-sided die is 1/12 when expressed as a fraction.

Cameron buys a bag of cookies that contains 6 chocolate chip cookies, 6 peanut butter cookies, 7 sugar cookies and 9 oatmeal cookies.

What is the probability that Cameron reaches in the bag and randomly selects an oatmeal cookie from the bag, eats it, then reaches back in the bag and randomly selects a peanut butter cookie?

$0.0714$0.0714​Correct  

Give your answer as a fraction, or accurate to at least 4 decimal places.

Probability of selecting an oatmeal cookie first:

There are 9 oatmeal cookies out of a total of 6 + 6 + 7 + 9 = 28 cookies in the bag.

Probability (oatmeal first) = 9 / 28

Probability of selecting a peanut butter cookie second:

After selecting the oatmeal cookie, there are now 28 - 1 = 27 cookies left in the bag.

There are 6 peanut butter cookies left in the bag.

Probability (peanut butter second) = 6 / 27

Now, calculate the overall probability by multiplying the probabilities of both events:

Probability (oatmeal first and peanut butter second) = Probability (oatmeal first) * Probability (peanut butter second)

Probability (oatmeal first and peanut butter second) = (9 / 28) * (6 / 27)

Now, calculate this product:

Probability (oatmeal first and peanut butter second) ≈ 0.0714

You are correct. The probability that Cameron randomly selects an oatmeal cookie first and then a peanut butter cookie from the bag is approximately 0.0714 when rounded to four decimal places. 

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	13	12	15	40
Female	11	19	8	38
Total	24	31	23	78

If one student is chosen at random,

Find the probability that the student was male AND got a "B".

Round answer 4 places after the decimal if needed.

$0.1538$0.1538​Correct  

To find the probability that a randomly chosen student was male AND got a "B," you can use the information from the table that summarizes the grades and gender of the students.

The number of male students who got a "B" is given as 12.

The total number of students is given as 78.

Now, you can find the probability using the following formula:

Probability (Male and B) = Number of Male Students who got a "B" / Total Number of Students

Probability (Male and B) = 12 / 78

Now, calculate the probability, rounding to four decimal places:

Probability (Male and B) ≈ 0.1538

So, the probability that a randomly chosen student was male AND got a "B" is approximately 0.1538 when rounded to four decimal places.

The table summarizes results from 985 pedestrian deaths that were caused by automobile accidents.

Driver Intoxicated?	Pedestrian Intoxicated?
	Yes	No
Yes	43	76
No	274	592

If one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was not intoxicated or the driver was intoxicated.

Report the answer as a percent rounded to one decimal place accuracy. You need not enter the "%" symbol.
prob = 72.2%

To find the probability that the pedestrian was not intoxicated or the driver was intoxicated, you can use the following formula:

Probability (P1 or P2) = Probability (P1) + Probability (P2) - Probability (P1 ∩ P2)

Let's calculate it:

Probability (P1):

Number of cases where the pedestrian was not intoxicated: 592 (from the "No" row in "Pedestrian Intoxicated?")

Total number of cases: 985

Probability (P1) = 592 / 985

Probability (P2):

Number of cases where the driver was intoxicated: 43 (from the "Yes" column in "Driver Intoxicated?")

Total number of cases: 985

Probability (P2) = 43 / 985

Probability (P1 ∩ P2):

Number of cases where the pedestrian was not intoxicated and the driver was intoxicated: 43 (from the intersection of "No" in "Pedestrian Intoxicated?" and "Yes" in "Driver Intoxicated?")

Total number of cases: 985

Probability (P1 ∩ P2) = 43 / 985

Now, calculate the probability of the pedestrian not being intoxicated or the driver being intoxicated:

Probability (P1 or P2) = Probability (P1) + Probability (P2) - Probability (P1 ∩ P2)

Probability (P1 or P2) = (592 / 985) + (43 / 985) - (43 / 985)

Probability (P1 or P2) ≈ 0.7222 (rounded to four decimal places)

To express this probability as a percent rounded to one decimal place:

Probability (P1 or P2) ≈ 72.2%

A test was given to a group of students. The grades and gender are summarized below

	A	B	C	Total
Male	15	16	17	48
Female	9	3	12	24
Total	24	19	29	72

If one student is chosen at random from those who took the test,

Find the probability that the student was male GIVEN they got an 'A'.

Round answer 4 places after the decimal if needed.

To find the probability that the student was male GIVEN they got an 'A,' you can use conditional probability:

P(Male | A) = P(Male and A)/ P(A)

Where:

P(Male | A) is the probability that the student was male given they got an 'A.'

P(Male and A) is the probability that the student was male and got an 'A.'

P(A)is the probability that a student got an 'A.'

First, let's calculate P(Male | A):

Number of males who got an 'A': 15

Total number of students who got an 'A': 24 (male + female)

P(Male | A)= 15/24

Now lets calculate P(A):

Total number of students who got an 'A': 24 (male + female)

Total number of students: 72

P(A)= 24/72

Now, we can find P(Male | A) using the formula:

P(Male | A) = P(Male | A) / P(A)

P(Male | A)= 15/24 / 24/72

Simplify:

P(Male | A) = 15/24 * 72/24

Now, calculate the value:

P(Male | A) = 15/24 * 72/24 = 0.625

So, the correct probability that the student was male given they got an 'A' is 0.625

A company estimates that 0.4% of their products will fail after the original warranty period but within 2 years of the purchase, with a replacement cost of $500.

If they offer a 2 year extended warranty for $47, what is the company's expected value of each warranty sold?

$$45$45​Correct  

Round 2 places after the decimal if needed.

To calculate the company's expected value for each warranty sold, we'll consider the probability of a product failing within 2 years after the original warranty period and the associated replacement cost.

Let's break down the calculation:

Probability of a product failing within 2 years after the original warranty period:

The probability is given as 0.4%, which can be expressed as a decimal by dividing by 100: 0.004.

Replacement cost if a product fails within 2 years:

The replacement cost is $500.

Cost of the extended warranty:

The cost of the extended warranty is $47.

Now, calculate the expected value (EV) of each warranty sold:

EV= (Probability of Failure x Replacement Cost) - Cost of Warranty

Cost of Warranty

EV=(Probability of Failure×Replacement Cost)−Cost of Warranty

EV = (0.004 \times $500) - $47

EV = ($2) - $47

EV = $45

So, the company's expected value of each warranty sold is $45. 

Use the spinner below.



P(9 or 1) = $0.1667$0.1667​Correct  

Round answer 4 places after the decimal if needed.

If the spinner has numbers from 1 to 12, and you want to find the probability of landing on either 9 or 1, you can calculate it as follows:

There are two favorable outcomes (landing on 9 or 1) out of a total of 12 possible outcomes (numbers 1 through 12 on the spinner).

So, the probability of landing on either 9 or 1 is:

P(9 or 1 ) = Number of Favorable Outcomes / Total number of possible outcomes = 2/12 = 1/6

Then divide 1/6 =0.1666666666666667

Round 4 decimal places: 0.1667

Suppose that 36% of people own dogs. If you pick two people at random, what is the probability that they both own a dog?

Give your answer as a decimal (to at least 3 places) or fraction

$0.1296$0.1296​Correct  

To find the probability that both people own a dog when picking two people at random, you can use the concept of independent events. The probability that the first person owns a dog is 36%, and the probability that the second person also owns a dog is also 36%, assuming that the events are independent.

To calculate the probability of both events happening, you multiply the probabilities of each event:

Probability(both own a dog) = Probability(person 1 owns a dog) * Probability(person 2 owns a dog)

Probability(both own a dog) = 0.36 * 0.36

Now, calculate the product:

Probability(both own a dog) = 0.36 * 0.36 = 0.1296

Suppose a jar contains 9 red marbles and 30 blue marbles. If you reach in the jar and pull out 2 marbles at random at the same time, find the probability that both are red. Answer should be in fractional form.

$\frac{72}{1482}$721482​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

To find the probability that both marbles drawn are red, you can use the following formula:

$�����������(���ℎ���)=\frac{���������������ℎ����2����������}{��������������������ℎ����2�������}$

First, let's calculate the total number of ways to choose 2 marbles from the jar without regard to their color. This can be calculated using combinations:

$��������������������ℎ����2�������=�(39,2)$

Now, let's calculate the number of ways to choose 2 red marbles from the 9 red marbles:

$���������������ℎ����2����������=�(9,2)$

Calculating the combinations:

$�(9,2)=36$ $�(39,2)=741$

Now, plug these values into the probability formula:

$�����������(���ℎ���)=\frac{36}{741}$

To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor (GCD), which is 3:

$�����������(���ℎ���)=\frac{36\mathrm{/}3}{741\mathrm{/}3}$

$�����������(���ℎ���)=\frac{12}{247}$

So, the correct probability that both marbles drawn are red is $12\mathrm{/}247$ in simplified fractional form, which can also be expressed as $72\mathrm{/}1482$

Suppose you roll a special 16-sided die. What is the probability that the number rolled is a "1" OR a "2"?

Round answer 4 places after the decimal if needed.

$0.125$0.125​Correct  

To find the probability of rolling either a "1" or a "2" on a 16-sided die, you need to consider the number of favorable outcomes (rolling a "1" or a "2") and the total number of possible outcomes (16 sides on the die).

Number of favorable outcomes (rolling a "1" or a "2") = 2

Total number of possible outcomes (16-sided die) = 16

Now, calculate the probability:

Probability(rolling a "1" or a "2") = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)

Probability(rolling a "1" or a "2") = 2 / 16

Simplify the fraction:

Probability(rolling a "1" or a "2") = 1 / 8

Rounded to four decimal places, the probability of rolling either a "1" or a "2" on a 16-sided die is approximately 0.1250.

6 cards are drawn at random from a standard deck.

Find the probability that all the cards are hearts. 

  

Find the probability that all the cards are face cards. 

 

Note: Face cards are kings, queens, and jacks.

Find the probability that all the cards are even. 

  

(Consider aces to be 1, jacks to be 11, queens to be 12, and kings to be 13)

To find the probabilities of drawing specific combinations of cards from a standard deck, you can use the principles of probability. Let's calculate each probability:

Probability that all the cards drawn are hearts:

There are 13 hearts in a standard deck.

There are 52 cards in a standard deck.

When drawing the first card, there are 13 hearts out of 52 cards.

After drawing the first heart, there are 12 hearts out of 51 cards for the second card.

Continue this process for all 6 cards.

Probability(all hearts) = (13/52) * (12/51) * (11/50) * (10/49) * (9/48) * (8/47)

To simplify this fraction, you can calculate the product of the numerators and the product of the denominators.

Probability(all hearts) = (13 * 12 * 11 * 10 * 9 * 8) / (52 * 51 * 50 * 49 * 48 * 47)

Probability that all the cards drawn are face cards (kings, queens, and jacks):

There are 12 face cards (4 kings, 4 queens, and 4 jacks) in a standard deck.

There are 52 cards in a standard deck.

When drawing the first card, there are 12 face cards out of 52 cards.

Continue this process for all 6 cards.

Probability(all face cards) = (12/52) * (11/51) * (10/50) * (9/49) * (8/48) * (7/47)

To simplify this fraction, you can calculate the product of the numerators and the product of the denominators.

Probability(all face cards) = (12 * 11 * 10 * 9 * 8 * 7) / (52 * 51 * 50 * 49 * 48 * 47)

Probability that all the cards drawn are even (considering aces as 1):

There are 20 even cards (2, 4, 6, 8, 10) in a standard deck (4 of each).

There are 52 cards in a standard deck.

When drawing the first card, there are 20 even cards out of 52 cards.

Continue this process for all 6 cards.

Probability(all even) = (20/52) * (19/51) * (18/50) * (17/49) * (16/48) * (15/47)

To simplify this fraction, you can calculate the product of the numerators and the product of the denominators.

Probability(all even) = (20 * 19 * 18 * 17 * 16 * 15) / (52 * 51 * 50 * 49 * 48 * 47)

Now, you can calculate these probabilities, which are expressed as unreduced fractions.

Probability that all the cards drawn are hearts:

There are 13 hearts in a standard deck.

There are 52 cards in a standard deck.

When drawing the first card, there are 13 hearts out of 52 cards.

After drawing the first heart, there are 12 hearts out of 51 cards for the second card.

Continue this process for all 6 cards.

Probability(all hearts) = (13/52) * (12/51) * (11/50) * (10/49) * (9/48) * (8/47)

To simplify this fraction, you can calculate the product of the numerators and the product of the denominators.

Probability(all hearts) = (13 * 12 * 11 * 10 * 9 * 8) / (52 * 51 * 50 * 49 * 48 * 47)

Probability(all hearts) = 1235520 / 14658134400

Probability that all the cards drawn are face cards (kings, queens, and jacks):

There are 12 face cards (4 kings, 4 queens, and 4 jacks) in a standard deck.

There are 52 cards in a standard deck.

When drawing the first card, there are 12 face cards out of 52 cards.

Continue this process for all 6 cards.

Probability(all face cards) = (12/52) * (11/51) * (10/50) * (9/49) * (8/48) * (7/47)

To simplify this fraction, you can calculate the product of the numerators and the product of the denominators.

Probability(all face cards) = (12 * 11 * 10 * 9 * 8 * 7) / (52 * 51 * 50 * 49 * 48 * 47)

Probability(all face cards) = 665280 / 14658134400

Probability that all the cards drawn are even (considering aces as 1):

There are 20 even cards (2, 4, 6, 8, 10) in a standard deck (4 of each).

There are 52 cards in a standard deck.

When drawing the first card, there are 20 even cards out of 52 cards.

Continue this process for all 6 cards.

Probability(all even) = (20/52) * (19/51) * (18/50) * (17/49) * (16/48) * (15/47)

To simplify this fraction, you can calculate the product of the numerators and the product of the denominators.

Probability(all even) = (20 * 19 * 18 * 17 * 16 * 15) / (52 * 51 * 50 * 49 * 48 * 47)

Probability(all even) = 96909120 / 14658134400

So, the probabilities for each scenario, expressed as fractions with the correct denominators, are as follows:

Probability that all the cards drawn are hearts: 1235520 / 14658134400

Probability that all the cards drawn are face cards: 665280 / 14658134400

Probability that all the cards drawn are even: 96909120 / 14658134400

A die is rolled twice. What is the probability of showing a 5 on the first roll and an even number on the second roll?

Your answer is : $\frac{3}{36}$336​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Answer should be in fractional form.

I apologize for any confusion. Let's recalculate:

To find the probability of rolling a 5 on the first roll and an even number on the second roll of a fair six-sided die, you can calculate it as follows:

Probability of rolling a 5 on the first roll:

There is 1 way to roll a 5 (rolling a 5).

There are 6 possible outcomes on the first roll.

Probability(5 on the first roll) = 1/6

Probability of rolling an even number on the second roll:

There are 3 even numbers on a die (2, 4, and 6).

There are 6 possible outcomes on the second roll.

Probability(even number on the second roll) = 3/6 = 1/2

Now, to find the probability of both events happening (rolling a 5 on the first roll and an even number on the second roll), multiply the probabilities of each event:

Probability(5 on the first roll and even number on the second roll) = (1/6) * (1/2) = 1/12

Non simplified = (36/3) 

A quick quiz consists of a multiple-choice question with 5 possible answers followed by a multiple-choice question with 4 possible answers. If both questions are answered with random guesses, find the probability that both responses are correct.

Report the answer as a percent rounded to one decimal place accuracy. You need not enter the "%" symbol.
prob = Correct%

To find the probability that both responses are correct when answered with random guesses:

Probability of getting the first question correct = 1/5 (There is 1 correct answer out of 5 choices).

Probability of getting the second question correct = 1/4 (There is 1 correct answer out of 4 choices).

Now, to find the probability that both events happen (both responses are correct), multiply the probabilities of each event:

Probability(both correct) = (1/5) * (1/4) = 1/20

To express this probability as a percent, multiply by 100:

Probability(both correct) = (1/20) * 100 = 5%

So, the probability that both responses are correct when answered with random guesses is 5%, rounded to one decimal place.

In a large population, 60 % of the people have been vaccinated. If 4 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

Give your answer as a decimal (to at least 3 places) or fraction. 

To find the probability that at least one of the four randomly selected people has been vaccinated, we can use the complement rule, which states that:

P(at least one) = 1 - P(None)

First, let's calculate the probability that none of the four people has been vaccinated:

Probability(none vaccinated) = (0.4)^4

This is because for each individual, the probability of not being vaccinated is 1 - 0.6 = 0.4, and we're assuming these events are independent.

Now, we can calculate the probability that at least one of them has been vaccinated using the complement rule:

Probability(at least one vaccinated) = 1 - Probability(none vaccinated)

Probability(at least one vaccinated) = 1 - (0.4)^4

Calculating:

Probability(at least one vaccinated) = 1 - 0.4^4 = 1 - 0.0256 = 0.9744

So, the probability that at least one of the four randomly selected people has been vaccinated is approximately 0.9744 (rounded to three decimal places).

Mrs. Donovan is the coach of a middle school golf team that has 10 golfers. There are 3 sixth graders, 3 seventh graders, and 4 eighth graders on the team. Mrs. Donovan will randomly select two team captains.

What is the probability that both captains will be eighth graders?

$\frac{2}{15}$215​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Make sure to write your answer as a reduced fraction.

To find the probability that both team captains will be eighth graders, you can use combinations.

First, let's calculate the total number of ways Mrs. Donovan can choose two team captains from the 10 golfers:

Total ways to choose two captains from 10 golfers = C(10, 2)

Now, let's calculate the number of ways to choose both captains from the 4 eighth graders:

Ways to choose 2 captains from 4 eighth graders = C(4, 2)

Now, you can find the probability:

Probability(both captains are eighth graders) = (Ways to choose 2 captains from 4 eighth graders) / (Total ways to choose two captains from 10 golfers)

Probability(both captains are eighth graders) = [C(4, 2)] / [C(10, 2)]

Now, calculate the combinations:

C(4, 2) = 6

C(10, 2) = 45

Now, plug these values into the probability formula:

Probability(both captains are eighth graders) = (6 / 45)

To simplify the fraction, you can divide both the numerator and denominator by their greatest common divisor (GCD), which is 3:

Probability(both captains are eighth graders) = (6 / 3) / (45 / 3)

Probability(both captains are eighth graders) = 2 / 15

So, the probability that both captains will be eighth graders is 2/15 in reduced fraction form.

A couple plans to have three children. What is the probability that

a) they have all girls? $\frac{1}{8}$18​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

b) they have at least one boy? $\frac{7}{8}$78​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Answers should be in fractional form.

Let's calculate the probabilities for each scenario:

a) Probability that they have all girls:

Each child's gender is independent, and the probability of having a girl is 1/2. To find the probability that all three children are girls, you multiply the probabilities together:

Probability(all girls) = (1/2) * (1/2) * (1/2) = (1/2)^3 = 1/8

So, the probability that they have all girls is 1/8.

b) Probability that they have at least one boy:

To find the probability that they have at least one boy, you can use the complement rule, which states:

Probability(at least one boy) = 1 - Probability(no boys)

To calculate the probability of having no boys (all girls), as we calculated in part (a):

Probability(no boys) = 1/8

Now, use the complement rule:

Probability(at least one boy) = 1 - 1/8 = 7/8

So, the probability that they have at least one boy is 7/8.

In a large population, 60% of the people have been vaccinated. If 5 people are randomly selected, what is the probability that at least one of them has been vaccinated?

Give your answer as a decimal to 4 places.

To find the probability that at least one of the five randomly selected people has been vaccinated, we can use the complement rule. The complement rule states that:

P(at least one)= 1 - P (none)

First, let's calculate the probability that none of the five people has been vaccinated:

Probability(none vaccinated) = (0.4)^5

This is because for each individual, the probability of not being vaccinated is 1 - 0.6 = 0.4, and we're assuming these events are independent.

Now, we can calculate the probability that at least one of them has been vaccinated using the complement rule:

Probability(at least one vaccinated) = 1 - Probability(none vaccinated)

Probability(at least one vaccinated) = 1 - (0.4)^5

Calculating:

Probability(at least one vaccinated) = 1 - 0.4^5 = 1 - 0.01024 ≈ 0.9898

So, the probability that at least one of the five randomly selected people has been vaccinated is approximately 0.9898 when rounded to four decimal places.

The spinner below is spun twice. If the spinner lands on a border, that spin does not count and spin again. It is equally likely that the spinner will land in each of the six sectors.

[Graphs generated by this script: initPicture(-100,100,-100,100);stroke='red'; fill='red';arc([90,0],[45,77.94228],90);path([[0,0],[90,0],[45,77.94228],[0,0]]);arc([-45,77.94228],[-90,0],90);path([[0,0],[-90,0],[-45,77.94228],[0,0]]);arc([-45,-77.94228],[45,-77.94228],90);path([[0,0],[-45,-77.94228],[45,-77.94228],[0,0]]);stroke='blue'; fill='blue';arc([45,77.94228],[-45,77.94228],90);path([[0,0],[45,77.94228],[-45,77.94228],[0,0]]);arc([-90,0],[-45,-77.94228],90);path([[0,0],[-90,0],[-45,-77.94228],[0,0]]);stroke='cyan'; fill='cyan';arc([45,-77.94228],[90,0],90);path([[0,0],[45,-77.94228],[90,0],[0,0]]);stroke='black'; fill='none';circle([0,0],90);line([45,77.94228],[-45,-77.94228]);line([-45,77.94228],[45,-77.94228]);line([-90,0],[90,0]);fontfamily='helvetica';fontfill='white';fontstyle='normal';fontweight='bold';text([55,25],'RED');text([-50,25],'RED');text([0,-70],'RED');text([0,65],'BLUE');text([-55,-30],'BLUE');fontfill='black';text([50,-25],'CYAN');strokewidth='5';line([-35,-35],[30,30]);line([25,15],[30,30]);line([15,25],[30,30]);]

For each question below, enter your response as a reduced fraction.

Find the probability of spinning cyan on the first spin and blue on the second spin.
$\frac{1}{18}$118​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Find the probability of spinning blue on the first spin and cyan on the second spin.
$\frac{1}{18}$118​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Find the probability of NOT spinning red on either spin. (Not red on the first spin and not red on the second spin.)

Find the probability of NOT spinning red on either spin. (Not red on the first spin and not red on the second spin.)

ansr: 1/4

1. From the figure, we get n(S) = 6 as there are six regions that the pointer can land on.

There are only one cyan region, so n (cyan on the first spin) = 1.
Thus, the probability of the pointer landing on cyan on the first spin is = 1/6
There are two blue regions, so n (blue on the second spin) = 2.
Thus, the probability of the pointer landing on blue on the second spin is = 2/6 = 1/3

Thus,the probability of spinning cyan on the first spin and blue on the second spin = 1/6 × 1/3 = 1/18

2. There are three red regions, so n (red on the first spin) = 3.
Thus, the probability of the pointer landing on red on the first spin is = 3/6 = 1/2

There are two blue regions, so n (blue on the second spin) = 2.
Thus, the probability of the pointer landing on blue on the second spin is = 2/6 = 1/3

Thus, the probability of spinning red on the first spin and blue on the second spin = 1/2× 1/3 = 1/6

3. Probability of non red in the first spin = 1 - 1/2 = 1/2

Probability of non red in the second spin = 1 - 1/2 = 1/2

Thus, the probability of NOT spinning red on both spins = 1/2× 1/2 = 1/4

A fair coin is tossed 5 times. Compute the probability of tossing 5 heads in a row.
$\frac{1}{32}$132​​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  
Enter your response as a reduced fraction.

To compute the probability of tossing 5 heads in a row with a fair coin, we can use the probability of a single coin toss, which is 1/2 (since there are two equally likely outcomes: heads and tails).

Since each coin toss is independent of the others, the probability of getting 5 heads in a row is calculated by multiplying the probabilities of getting heads on each of the 5 tosses:

Probability(5 heads in a row) = (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = (1/2)^5 = 1/32

So, the probability of tossing 5 heads in a row is 1/32 when expressed as a reduced fraction.

A jar contains 8 red marbles numbered 1 to 8 and 12 blue marbles numbered 1 to 12. A marble is drawn at random from the jar. Find the probability of the given event, please show your answers as reduced fractions.

(a) The marble is red. P(red)= $\frac{2}{5}$25​​Correct   ${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$

(b) The marble is odd-numbered. P(odd)= $\frac{1}{2}$12​​Correct   ${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$

(c) The marble is red or odd-numbered. P(red or odd) =7/10

(d) The marble is blue or even-numbered. P(blue or even) =4/5

Let's calculate the probabilities for each event:

(a) The marble is red.

There are 8 red marbles out of a total of 8 red + 12 blue marbles.

P(red) = 8/20 = 2/5

(b) The marble is odd-numbered.

There are 4 odd-numbered red marbles (1, 3, 5, 7) and 6 odd-numbered blue marbles (1, 3, 5, 7, 9, 11) out of a total of 8 red + 12 blue marbles.

P(odd) = (4 + 6) / 20 = 10/20 = 1/2

(c) The marble is red or odd-numbered.

To calculate this probability, we can use the principle of inclusion-exclusion. First, find the probability of a red marble and the probability of an odd-numbered marble separately, and then subtract the probability of both events happening at the same time.

P(red or odd) = P(red) + P(odd) - P(red and odd)

P(red and odd) = (4/20) # The odd-numbered red marbles (1, 3, 5, 7)

P(red or odd) = (2/5) + (1/2) - (4/20) = (4/10) + (5/10) - (4/20) = (9/10) - (4/20) = (9/10) - (1/5) = (9/10) - (2/10) = 7/10

(d) The marble is blue or even-numbered.

Similar to part (c), we can calculate this probability using the inclusion-exclusion principle.

P(blue or even) = P(blue) + P(even) - P(blue and even)

P(blue and even) = (6/20) # The even-numbered blue marbles (2, 4, 6, 8, 10, 12)

P(blue or even) = (12/20) + (4/20) - (6/20) = (3/5) + (1/5) - (3/10) = (4/5) - (3/10) = (8/10) - (3/10) = 5/10 = 1/2

So, the probabilities are:

(a) P(red) = 2/5

(b) P(odd) = 1/2

(c) P(red or odd) = 7/10

(d) P(blue or even) = 1/2

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	6	3	18	27
Female	10	11	12	33
Total	16	14	30	60

If one student is chosen at random,

Find the probability that the student got a B: $0.2333$0.2333​Correct  

Find the probability that the student was female AND got a "C": $0.2$0.2​Correct  

Find the probability that the student was male OR got an "B": $0.6333$0.6333​Correct  

If one student is chosen at random, find the probability that the student got a 'C' GIVEN they are female: $0.3636$0.3636​Correct  

Round answers 4 places after the decimal if needed.

Let's calculate the probabilities for each event:

Probability that the student got a B:

There are 14 students who got a B out of a total of 60 students.

P(B) = 14/60 = 7/30

Probability that the student was female AND got a C:

There are 12 female students who got a C out of a total of 60 students.

P(Female and C) = 12/60 = 1/5

Probability that the student was male OR got a B:

To calculate this probability, we'll find the probability of being male and the probability of getting a B separately and then add them, subtracting the probability of being both male and getting a B (since it's counted twice).

P(Male or B) = P(Male) + P(B) - P(Male and B)

P(Male) = 27/60 # Number of male students out of the total

P(Male and B) = 3/60 # Number of male students who got a B out of the total

P(Male or B) = (27/60) + (7/30) - (3/60) = (27/60) + (14/60) - (3/60) = (41/60) - (3/60) = 38/60 = 19/30

Probability that the student got a 'C' GIVEN they are female:

To calculate this conditional probability, we'll use the formula:

P(C | Female) = P(Female and C) / P(Female)

P(Female) = 33/60 # Number of female students out of the total

P(C | Female) = (1/5) / (33/60) = (1/5) / (11/20) = (1/5) * (20/11) = (4/11)

Now, let's round the answers to 4 decimal places:

P(B) ≈ 0.2333

P(Female and C) = 0.2000

P(Male or B) = 0.6333

P(C | Female) ≈ 0.3636

A card is drawn randomly from a standard 52-card deck. Find the probability of the given event.
(a) The card drawn is 7
The probability is : $0.0769$0.0769​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

(b) The card drawn is a face card (Jack, Queen, or King)
The probability is : $0.2308$0.2308​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

(c) The card drawn is not a face card.
The probability is : $0.7692$0.7692​Correct${\displaystyle \frac{\mathrm{}}{\mathrm{}}}$  

Answers should be in Decimal form.

Let's calculate the probabilities for each event:

(a) The card drawn is 7:

There are four 7s in a standard 52-card deck (one 7 in each suit).

Probability(7) = (Number of 7s) / (Total number of cards) = 4/52 = 1/13

(b) The card drawn is a face card (Jack, Queen, or King):

In each of the four suits, there is one Jack, one Queen, and one King, making a total of 3 face cards per suit. There are four suits in a deck.

Probability(Face card) = (Number of face cards) / (Total number of cards) = (3 * 4) / 52 = 12/52 = 3/13

(c) The card drawn is not a face card:

There are 4 suits, and each suit has 13 cards. Out of those, 3 are face cards (Jack, Queen, King). So, there are 10 non-face cards in each suit.

Probability(Not a face card) = (Number of non-face cards) / (Total number of cards) = (10 * 4) / 52 = 40/52 = 10/13

So, the probabilities are:

(a) The card drawn is 7: 1/13

(b) The card drawn is a face card: 3/13

(c) The card drawn is not a face card: 10/13

Fraction to decimal 

1/13 =0.0769

3/13=0.2308

10/13=0.7692

Suppose a jar contains 13 red marbles and 16 blue marbles. If you reach in the jar and pull out 2 marbles at random at the same time, find the probability that both are red.

Answer should be in fractional form.

To find the probability that both marbles drawn are red, you can use the following steps:

Calculate the probability of drawing a red marble on the first draw:

There are 13 red marbles out of a total of 29 marbles in the jar.

Probability(First marble is red) = (Number of red marbles) / (Total number of marbles) = 13/29

Calculate the probability of drawing a red marble on the second draw, assuming that the first marble was red:

After the first draw, there are now 12 red marbles left out of a total of 28 marbles (since one red marble was already drawn).

Probability(Second marble is red | First marble is red) = (Number of remaining red marbles) / (Total number of remaining marbles) = 12/28

Now, to find the probability that both marbles are red, multiply the probabilities from step 1 and step 2:

Probability(Both marbles are red) = Probability(First marble is red) * Probability(Second marble is red | First marble is red)

Probability(Both marbles are red) = (13/29) * (12/28)

Simplify the fraction:

Probability(Both marbles are red) = (13/29) * (3/7)

Multiply the numerators and denominators:

Probability(Both marbles are red) = (13 * 3) / (29 * 7)

Probability(Both marbles are red) = 39/203

So, the probability that both marbles drawn are red is 39/203 when expressed as a reduced fraction.

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	12	15	2	29
Female	8	13	6	27
Total	20	28	8	56

If one student is chosen at random,

Find the probability that the student was female OR got an "C".

Round answer 4 places after the decimal if needed.

Answ:0.5179

To find the probability that the student was female OR got a "C," we can use the principle of inclusion-exclusion. First, find the probability of being female and the probability of getting a "C" separately, and then add them, subtracting the probability of being both female and getting a "C" (since it's counted twice).

Let's calculate each part:

Probability of being female:

There are 27 female students out of a total of 56 students.

P(Female) = 27/56

Probability of getting a "C":

There are 8 students who got a "C" out of a total of 56 students.

P(C) = 8/56

Probability of being both female and getting a "C":

There are 6 female students who got a "C" out of a total of 56 students.

P(Female and C) = 6/56

Now, let's calculate the probability of being female OR getting a "C":

P(Female OR C) = P(Female) + P(C) - P(Female and C)

P(Female OR C) = (27/56) + (8/56) - (6/56)

P(Female OR C) = (35/56) - (6/56)

P(Female OR C) = 29/56

Now, round the answer to 4 decimal places:

P(Female OR C) ≈ 0.5179

So, the probability that the student was female OR got a "C" is approximately 0.5179 when rounded to 4 decimal places.

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	20	12	10	42
Female	8	3	16	27
Total	28	15	26	69

If one student is chosen at random,

Find the probability that the student was female AND got a "A".

Round answer 4 places after the decimal if needed.

$0.1159$0.1159​Correct  

To find the probability that the student was female AND got an "A," you can use the following steps:

Calculate the probability of being female:

There are 27 female students out of a total of 69 students.

P(Female) = 27/69

Calculate the probability of getting an "A":

There are 28 students who got an "A" out of a total of 69 students.

P(A) = 28/69

Calculate the probability of being both female and getting an "A":

There are 8 female students who got an "A" out of a total of 69 students.

P(Female and A) = 8/69

Now, let's calculate the probability of being female AND getting an "A":

P(Female AND A) = P(Female) * P(A | Female)

P(A | Female) represents the probability of getting an "A" given that the student is female. To calculate this conditional probability, you can use the following formula:

P(A | Female) = (Number of females who got an "A") / (Total number of females)

P(A | Female) = (8/27)

Now, calculate P(Female AND A):

P(Female AND A) = (27/69) * (8/27)

Simplify the fraction:

P(Female AND A) = (1/3) * (8/27)

P(Female AND A) = (8/81)

Now, round the answer to 4 decimal places 0.1159

A quick quiz consists of a multiple-choice question with 3 possible answers followed by a multiple-choice question with 3 possible answers. If both questions are answered with random guesses, find the probability that both responses are correct.

Report the answer as a percent rounded to one decimal place accuracy. You need not enter the "%" symbol.
prob = Correct%

In a multiple-choice question with 3 possible answers, the probability of guessing the correct answer by random chance is 1/3.

Since both questions are answered with random guesses, the probability of both responses being correct is (1/3) * (1/3) = 1/9.

To report this as a percentage rounded to one decimal place, we can convert 1/9 to a percentage:

(1/9) * 100 = 11.11%

So, the probability that both responses are correct is approximately 11.11% when rounded to one decimal place.

A test was given to a group of students. The grades and gender are summarized below

	A	B	C	Total
Male	7	20	15	42
Female	10	4	11	25
Total	17	24	26	67

If one student is chosen at random from those who took the test,

Find the probability that the student was male GIVEN they got a 'C'. Round answer 4 places after the decimal if needed.

To find the probability that the student was male given they got a 'C', you can use the conditional probability formula:

$�(����\mathrm{\mid }�)=\frac{�(����\cap �)}{�(�)}$

Where:

• $�(����\mathrm{\mid }�)$ is the probability that the student was male given they got a 'C'.
• $�(����\cap �)$ is the probability that the student was male and got a 'C'.
• $�(�)$ is the probability that a student got a 'C'.

From the table, you can see that there were 15 males who got a 'C' (Male, C = 15), and the total number of students who got a 'C' is 26 (Total, C = 26).

Now, calculate the probability:

$�(����\mathrm{\mid }�)=\frac{15}{26}$

Rounded to 4 decimal places, the probability that the student was male given they got a 'C' is approximately 0.5769.

In a large population, 70 % of the people have been vaccinated. If 3 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

Give your answer as a decimal (to at least 3 places) or fraction. $0.973$0.973​Correct 

To find the probability that at least one of the three randomly selected people has been vaccinated, you can use the complement rule. First, calculate the probability that none of them has been vaccinated (complement of at least one), and then subtract that from 1.

Let's calculate the probability that none of them has been vaccinated:

Probability that one person is not vaccinated = 30% (since 100% - 70% = 30% are not vaccinated).

Probability that all three people are not vaccinated (assuming independence) = (0.3)^3 = 0.027.

Now, calculate the probability that at least one of them has been vaccinated:

Probability(at least one vaccinated) = 1 - Probability(none vaccinated) = 1 - 0.027 = 0.973.

So, the probability that at least one of the three randomly selected people has been vaccinated is approximately 0.973 when rounded to three decimal places.

Suppose that 21% of people own dogs. If you pick two people at random, what is the probability that they both own a dog?

Give your answer as a decimal (to at least 3 places) or fraction

$0.044$0.044​Correct  

To find the probability that both of the two randomly selected people own a dog, you can use the probability of owning a dog for each person and multiply them together.

Probability that one person owns a dog = 21% = 0.21 (in decimal form).

Since the events are independent (the probability of one person owning a dog doesn't affect the probability of the other person owning a dog), you can multiply these probabilities:

Probability(both own a dog) = (0.21) * (0.21) = 0.0441.

So, the probability that both of the two randomly selected people own a dog is approximately 0.0441 when rounded to three decimal places.

Project # 3 – Congress(B)

HOUSE OF REPRESENTATIVES

Political Party
	Republican	Democrat	Independent	Total
Male	170	192	0	362
Female	23	57	0	80
Total	193	249	0	442

 

SENATE

Political Party
	Republican	Democrat	Independent	Total
Male	38	50	1	89
Female	11	24	1	36
Total	49	74	2	125

 

COMBINED TABLE – CONGRESS - COMBINING HOUSE OF REPRESENTATIVES AND SENATE

Political Party
	Republican	Democrat	Independent	Total
Male	208	242	1	451
Female	34	81	1	116
Total	242	323	2	567

 

 

These charts are for exercise purposes only. It does not depict the actual numbers represented in the US Government today.

Must show calculations/formula for credit. Any answer given with no calculations shown will result in no credit for that answer.

 

Please answer the questions below and show your work where applicable. All answers should be proportions rounded to the nearest thousandths (3 decimals). For Example: this is what is required to be shown if I ask this question – shown is how you are expected to answer: (please note that this is the correct formula and process whenever an or question is asked)

Find the probability that a Senate is a Male or a Democrat:

 

 

 

1.       Find the probability that a randomly selected House of Representative is a male.

 

 

 

 

2.       Find the probability that a randomly selected Senator is a female.

 

 

 

 

 

3.       A House of Representative member is selected at random. Find the probability of each event:

a.       The representative is a female

 

 

 

b.       The representative is an independent.

 

 

 

 

c.       The representative is a female given that the representative is a Democrat

 

d.       The representative is female and a Democrat

 

 

 

4.       A Senator is selected at random. Find the probability of each event

a.       The senator is a female

 

 

 

 

b.       The senator is not a Republican

 

 

 

 

c.       The senator is female or a Democrat

   

 

 

d.       The senator is a male or a Democrat

 

 

 

5.       Using the same row and column headings, create a combined table for the Congress (template created for you above – you can just fill in the table I already created)

 

6.       A member of Congress is selected at random. Use the combined table to find the probability of each event.

 

a.       The member is Independent

 

b.       The member is female and a Democrat

 

c.       The member is male or a Democrat

 

Find the probability that a randomly selected House of Representative is a male.

To find this probability, you need to divide the number of male representatives by the total number of representatives in the House.

Probability(male representative) = Number of male representatives / Total number of representatives in the House

Probability(male representative) = 362 / 442 ≈ 0.818 (rounded to 3 decimals)

Find the probability that a randomly selected Senator is a female.

To find this probability, you need to divide the number of female senators by the total number of senators.

Probability(female senator) = Number of female senators / Total number of senators

Probability(female senator) = 36 / 125 ≈ 0.288 (rounded to 3 decimals)

A House of Representative member is selected at random. Find the probability of each event:

a. The representative is a female.

Probability(female representative) = Number of female representatives / Total number of representatives in the House

Probability(female representative) = 80 / 442 ≈ 0.181 (rounded to 3 decimals)

b. The representative is an independent.

Probability(independent representative) = Number of independent representatives / Total number of representatives in the House

Probability(independent representative) = 0 / 442 = 0 (since there are no independent representatives)

c. The representative is a female given that the representative is a Democrat.

This is a conditional probability. To find it, we need to calculate:

Probability(female representative | Democrat) = (Number of female representatives who are Democrats) / (Number of representatives who are Democrats)

Probability(female representative | Democrat) = 57 / 249 ≈ 0.229 (rounded to 3 decimals)

d. The representative is female and a Democrat.

Probability(female representative and Democrat) = Number of female representatives who are Democrats / Total number of representatives in the House

Probability(female representative and Democrat) = 57 / 442 ≈ 0.129 (rounded to 3 decimals)

A Senator is selected at random. Find the probability of each event:

a. The senator is a female.

Probability(female senator) = Number of female senators / Total number of senators

Probability(female senator) = 36 / 125 ≈ 0.288 (rounded to 3 decimals)

b. The senator is not a Republican.

Probability(not a Republican senator) = 1 - Probability(Republican senator)

Probability(not a Republican senator) = 1 - (Number of Republican senators / Total number of senators)

Probability(not a Republican senator) = 1 - (49 / 125) ≈ 0.608 (rounded to 3 decimals)

c. The senator is female or a Democrat.

Probability(female senator or Democrat) = Probability(female senator) + Probability(Democrat senator) - Probability(female senator and Democrat senator)

Probability(female senator or Democrat senator) = (36 / 125) + (74 / 125) - (24 / 125) ≈ 0.552 (rounded to 3 decimals)

d. The senator is a male or a Democrat.

Probability(male senator or Democrat senator) = Probability(male senator) + Probability(Democrat senator) - Probability(male senator and Democrat senator)

Probability(male senator or Democrat senator) = (89 / 125) + (74 / 125) - (24 / 125) ≈ 0.792 (rounded to 3 decimals)

Using the same row and column headings, create a combined table for the Congress.

Here's the combined table for the Congress:

Political Party | Republican | Democrat | Independent | Total

Male | 208 | 242 | 1 | 451

Female | 34 | 81 | 1 | 116

Total | 242 | 323 | 2 | 567

A member of Congress is selected at random. Use the combined table to find the probability of each event.

a. The member is Independent.

Probability(Independent member) = Number of independent members / Total number of members

Probability(Independent member) = 2 / 567 ≈ 0.004 (rounded to 3 decimals)

b. The member is female and a Democrat.

Probability(female member and Democrat) = Number of female members who are Democrats / Total number of members

Probability(female member and Democrat) = 81 / 567 ≈ 0.143 (rounded to 3 decimals)

c. The member is male or a Democrat.

Probability(male member or Democrat) = Probability(male member) + Probability(Democrat member) - Probability(male member and Democrat member)

Probability(male member or Democrat member) = (451 / 567) + (323 / 567) - (242 / 567) ≈ 0.896 (rounded to 3 decimals)