Use the table to answer the question.
Plant Types A B C D E
Number of Seedlings 100 140 80 60 40
A farmer conducted an inventory on his two square kilometers of rainforest; the results are in the table. He plans on randomly selecting a seedling to experiment on when he goes back to the forest. What is the probability that he will randomly pick a type B plant? Write your answer in the simplest form of a fraction.
(1 point)
The total number of seedlings is 420 (100+140+80+60+40). The number of type B seedlings is 140.
Therefore, the probability of picking a type B plant is:
140/420
Simplifying this fraction by dividing both the numerator and denominator by 140 (the greatest common factor), we get:
1/3
Therefore, the probability of picking a type B plant is 1/3 or approximately 0.333.
The letter tiles C, A, R, E, and S are placed in a box. Without looking, Jorelle picks a letter tile from the box and records the result before placing it back. He repeats the procedure 100 times and observes the following results:
Outcomes C A R E S
Frequency 20 25 30 10 15
Based on the observed frequency, what is the probability that Jorelle will pick a letter R? The answer should be in percent form.
(1 point)
The frequency of letter R is 30, out of a total of 100 picks.
Therefore, the probability of picking a letter R is:
30/100 = 0.3
To convert this decimal to a percentage, we multiply by 100:
0.3 x 100 = 30%
Therefore, the probability of picking a letter R is 30%.
Use the table to answer the question.
Outcomes 2 4 6 8 10 12
Frequency 10 9 6 15 13 8
Jede tossed a cube with faces numbered with 2, 4, 6, 8, 10, and 12. The results are recorded in the table. What is the largest discrepancy between the experimental and the expected probability of this experiment? The answer needs to be in percent form to the nearest whole number.
(1 point)
There are six possible outcomes and each outcome has a probability of 1/6 if the cube is fair.
Therefore, the expected probability for each outcome is:
1/6 = 0.1667 (rounded to four decimal places)
To calculate the experimental probability for each outcome, we divide the frequency by the total number of trials:
Experimental probability = frequency/total trials
For example, the experimental probability of getting a 2 is:
10 / (10+9+6+15+13+8) = 0.1923 (rounded to four decimal places)
We can calculate the experimental probability for each outcome using the same formula.
To find the largest discrepancy between the experimental and expected probability, we need to find the absolute difference between the two for each outcome and then find the largest value.
For example, the absolute difference for getting a 2 is:
|0.1923 - 0.1667| = 0.0256 (rounded to four decimal places)
We can calculate the absolute difference for each outcome using the same formula.
The largest discrepancy is between the experimental probability of getting an 8 and the expected probability.
The expected probability of getting an 8 is 0.1667.
The experimental probability of getting an 8 is:
15 / (10+9+6+15+13+8) = 0.2885 (rounded to four decimal places)
The absolute difference is:
|0.2885 - 0.1667| = 0.1218 (rounded to four decimal places)
Therefore, the largest discrepancy between the experimental and the expected probability is 12% (rounded to the nearest whole number).
Use the table to answer the question.
Outcomes K I N D
Frequency 120 140 105 135
A spinner is divided into 4 sections labeled as K, I, N, D. Xavier reproduced the wheel and uses a computer to simulate the outcomes of 500 spins. What is the approximate probability that the spinner will stop on a consonant on the next spin?
(1 point)
There are two consonants on the spinner, K and N. The total number of outcomes in one spin is 4.
The probability that the spinner will stop on a consonant in one spin is:
P(consonant) = number of consonants / total number of outcomes
P(consonant) = 2/4 = 0.5
This means that the probability of getting a consonant on one spin is 0.5 or 50%.
Since Xavier simulated 500 spins, we can estimate the number of spins that will result in a consonant by multiplying the probability by the total number of spins:
Number of spins with a consonant = P(consonant) x Total number of spins
Number of spins with a consonant = 0.5 x 500
Number of spins with a consonant = 250
Therefore, the approximate probability that the spinner will stop on a consonant on the next spin is 250/500, or 0.5 (or 50%)
The letter tiles C, A, R, E, and S are placed in a box. Without looking, Jorelle picks a letter tile from the box and records the result before placing it back. He repeats the procedure 100 times and observes the following results:
Outcomes C A R E S
Frequency 20 25 30 10 15
Based on the observed frequency, develop a probability model for this experiment. Express the probability in decimal form, rounded to the nearest hundredth.
(1 point)
The probability of selecting each letter can be calculated by dividing the frequency of that letter by the total number of trials, which is 100 in this case.
The probability of selecting C is:
P(C) = 20/100 = 0.20
The probability of selecting A is:
P(A) = 25/100 = 0.25
The probability of selecting R is:
P(R) = 30/100 = 0.30
The probability of selecting E is:
P(E) = 10/100 = 0.10
The probability of selecting S is:
P(S) = 15/100 = 0.15
So the probability model for this experiment is:
P(C) = 0.20
P(A) = 0.25
P(R) = 0.30
P(E) = 0.10
P(S) = 0.15
Probability of Chance Events Quick Check
4 of 54 of 5 Items
Question
Use the tables to answer the question.
Simon’s Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A 5 15
Bowl B 16 4
Clark’s Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Simon and Clark repeatedly selected a ball from both bowls and recorded the results in a table. Whose results will give you a better indication about the proportion of white and red balls in each bowl? Explain your answer.
(1 point)
Clark's results will give a better indication about the proportion of white and red balls in each bowl.
The reason for this is the sample size. Clark has selected balls many more times than Simon, so Clark's results are more representative of the overall proportion of white and red balls in each bowl.
For example, in Bowl B, Clark has selected a total of 100 balls (60 white, 40 red), while Simon has only selected 20 balls in total (16 white, 4 red). Clark's sample is five times larger than Simon's sample, so Clark's results are likely to be more accurate and representative of the true proportion of white and red balls in the bowl.
In general, the larger the sample size, the more accurate and representative the results will be.
Use the table to answer the question.
Number of White Balls Selected Number of Red Balls Selected
Bowl A 5 15
Bowl B 16 4
Bowls A and B contain a number of white and red balls. Simon repeatedly selected a ball from both bowls and recorded the results in a table. Using Simon’s results, what is the experimental probability of picking a red ball from Bowl A? The answer needs to be in percent form.
(1 point)
In Bowl A, Simon selected 15 red balls out of a total of 20 balls (5 white + 15 red).
The experimental probability of picking a red ball from Bowl A is:
Number of red balls selected / Total number of balls selected
= 15 / 20
= 0.75
To express this probability as a percentage, we multiply by 100:
0.75 x 100
= 75%
Therefore, the experimental probability of picking a red ball from Bowl A, based on Simon's results, is 75%.
A basket of fruits contains 5 apples and 3 pears. Sharon took two fruits at random. What is the probability that both fruits are apples? Write your answer in the simplest form of fraction.(1 point)
The probability of getting an apple on the first draw is 5/8, since there are 5 apples out of 8 total fruits.
After the first apple is drawn, there are only 4 apples and 7 total fruits left. Therefore, the probability of drawing a second apple is 4/7.
To find the probability of both events happening, we multiply the probabilities:
P(both apples) = (5/8) x (4/7)
Simplifying the fraction by canceling out common factors (2 and 4 in the numerator and 4 in the denominator), we get:
P(both apples) = (5/8) x (1/7)
P(both apples) = 5/56
Therefore, the probability that both fruits are apples is 5/56.
A coin is flipped and a spinner is spun simultaneously. The spinner is divided into six equally sized sections labeled with 1, 2, 3, 4, 5, 6. What is the probability of flipping a tail and landing on 5?(1 point)
The probability of flipping a tail is 1/2 and the probability of landing on 5 on the spinner is 1/6.
To find the probability of both events happening, we multiply the probabilities:
P(tail and landing on 5) = (1/2) x (1/6)
Simplifying the fraction, we get:
P(tail and landing on 5) = 1/12
Therefore, the probability of flipping a tail and landing on 5 is 1/12.
Use the table to answer the question.
Studying Online
Girls Boys Row total
Want the option to keep studying online 10 3 13
Don’t want the option to keep studying online 5 7 12
Column total 15 10 25What is the probability of selecting a student who wants the option to keep studying online? Write your answer in percent form.
The table shows a total of 13 students who want the option to keep studying online out of a total of 25 students:
Girls Boys Row total
Want the option to keep studying online 10 3 13
Don’t want the option to keep studying online 5 7 12
Column total 15 10 25
The probability of selecting a student who wants the option to keep studying online is:
Number of students who want to keep studying online / Total number of students
= 13/25
To express this probability as a percentage, we multiply by 100:
(13/25) x 100 = 52%
Therefore, the probability of selecting a student who wants the option to keep studying online is 52%.
A money envelope contains five $10 bills, ten $20 bills, and fifteen $50 bills. Sandra randomly selects two bills without returning the first bill. What is the probability of getting a $10 bill then a $50 bill? Write your answer in the simplest form of fraction.(1 point)
The probability of selecting a $10 bill on the first draw is 5/30, since there are 5 $10 bills out of 30 total bills.
After the first bill is drawn, there are 29 total bills left, including 4 $10 bills and 15 $50 bills. Therefore, the probability of drawing a $50 bill second is 15/29.
To find the probability of both events happening, we multiply the probabilities:
P($10 and $50) = (5/30) x (15/29)
Simplifying the fraction by canceling out the common factor of 5 in the numerator and denominator of the first term, we get:
P($10 and $50) = (1/6) x (15/29)
Simplifying the fraction further by canceling out common factors (3 in the numerator and denominator of the second term), we get:
P($10 and $50) = (1/2) x (5/29)
P($10 and $50) = 5/58
Therefore, the probability of getting a $10 bill then a $50 bill is 5/58.
A spinner labeled with the letters of the word HONEST is to be spun once. Describe the probability of landing on an unshaded section in the sample space.
(1 point)
There is no information given in the question about which sections of the spinner are unshaded or how many sections are unshaded. Therefore, we cannot accurately describe the probability of landing on an unshaded section without this information.
However, we can say that the probability of landing on any section of the spinner is 1/6, since there are six letters in the word HONEST and each letter corresponds to one equally sized section on the spinner.
Which set of events is dependent?(1 point)
Responses
choosing a marble from a box, replacing it, then choosing another marble
the event of getting two heads when flipping two fair coins
choosing a ball from a bag and then choosing another ball without replacing the first
rolling a number cube and flipping a coin
The set of events that is dependent is:
choosing a ball from a bag and then choosing another ball without replacing the first.
This is because the second event is affected by the outcome of the first event. If the first ball is removed from the bag, the total number of balls in the bag is reduced and the probability of selecting a particular ball on the second draw changes. Therefore, the outcome of the first event affects the outcome of the second event, making the events dependent.
A basket of fruits contains 5 bananas, 8 mangoes, and 7 apples. Two fruits are randomly selected without replacement. Calculate the probability of selecting an apple and a banana.(1 point)
The probability of selecting an apple first is 7/20, since there are 7 apples out of 20 total fruits.
After the first fruit is removed, the probability of selecting a banana is 5/19, since there are 5 bananas left out of 19 total fruits.
To find the probability of both events happening, we multiply the probabilities:
P(apple and banana) = (7/20) x (5/19)
Simplifying the fraction by canceling out the common factor of 5 in the numerator and denominator of the first term, we get:
P(apple and banana) = (7/20) x (1/19)
Simplifying the fraction further by canceling out common factors (1 in the numerator and 4 in the denominator of the first term), we get:
P(apple and banana) = (7/20) x (1/19)
P(apple and banana) = 7/380
Therefore, the probability of selecting an apple and a banana is 7/380.
A basket of fruits contains 10 bananas, 5 mangoes, and 5 apples. A fruit is selected at random. Calculate the probability of selecting an apple or a banana. Express the answer in percent form.(1 point)
Responses
The probability of selecting an apple is 5/20 (i.e., 5 apples out of 20 fruits) and the probability of selecting a banana is 10/20 (i.e., 10 bananas out of 20 fruits).
To calculate the probability of selecting an apple or a banana, we add the probabilities of selecting an apple and selecting a banana, since these events are mutually exclusive (i.e., selecting an apple excludes selecting a banana, and vice versa):
P(apple or banana) = P(apple) + P(banana)
P(apple or banana) = (5/20) + (10/20)
P(apple or banana) = 15/20 = 0.75
To express this probability as a percentage, we multiply by 100:
0.75 x 100 = 75%
Therefore, the probability of selecting an apple or a banana is 75%.
Use the table to answer the question.
Studying Online
Girls Boys Row total
Want the option to keep studying online 10 3 13
Don’t want the option to keep studying online 5 7 12
Column total 15 10 25
Which mathematical sentence can be used to calculate the probability of selecting a student who wants the option to keep studying online or is a boy?
(1 point)
To calculate the probability of selecting a student who wants the option to keep studying online or is a boy, we can use the formula for the probability of a union of two events:
P(A or B) = P(A) + P(B) - P(A and B)
where A and B represent two distinct events.
In this case, we can let A represent the event of selecting a student who wants the option to keep studying online, and let B represent the event of selecting a boy.
Using the data from the table, we can calculate the probabilities of each event individually:
P(A) = 13/25
P(B) = 10/25
To find the probability of both events happening, we need to determine the probability of selecting a student who wants the option to keep studying online and is a boy. From the table, we can see