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?
The probability of flipping a tail is 1/2 and the probability of landing on 5 on the spinner is 1/6. Since the events are happening simultaneously, we can use the Multiplication Rule of Probability, which states that the probability of two independent events happening together is the product of their individual probabilities. Therefore, the probability of flipping a tail and landing on 5 is:
1/2 x 1/6 = 1/12
So the probability of flipping a tail and landing on 5 is 1/12 or approximately 0.083.
Since events A and B are independent, the probability of both occurring is the product of their individual probabilities:
P(A and B) = P(A) x P(B) = 0.3 x 0.25 = 0.075
Therefore, the probability of events A and B happening at the same time is 0.075 or 7.5%.
The total number of students who want the option to keep studying online is 13, out of a total of 25 students. Therefore, the probability of selecting a student who wants the option to keep studying online is:
13/25 = 0.52
Multiplying by 100% to convert to percent form gives:
0.52 x 100% = 52%
Therefore, the probability of selecting a student who wants the option to keep studying online is 52%.
The probability of getting a $10 bill on the first draw is 5/30 (since there are 5 $10 bills out of 30 total bills). After this bill is not returned, there are 29 bills left in the envelope, of which 15 are $50 bills. Therefore, the probability of getting a $50 bill on the second draw, given that a $10 bill was drawn first and not returned, is 15/29.
Since we need both events to happen (getting a $10 bill then a $50 bill), we can use the multiplication rule and multiply the probabilities of the two events:
P($10, then $50) = (5/30) x (15/29)
Simplifying this fraction by canceling out common factors gives:
P($10, then $50) = (1/6) x (15/29) = 5/58
Therefore, the probability of getting a $10 bill then a $50 bill is 5/58 in the simplest form of fraction.
If two events are independent, the probability of both events occurring is the product of their individual probability. Events A and B are independent with P(A)=0.3 and P(B)=0.25. What is the probability of events A and B happening at the same time? Write your answer in decimal form.
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.
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
H - shaded
O - not shaded
N - not shaded
E - shaded
S - shaded
T - not shaded
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.
A: 1/6
B: 3
C: 1/3
D:1/2
There are three unshaded sections on the spinner, labeled O, N, and T. Since there are a total of six sections on the spinner, each labeled with one of the letters in the word HONEST, the probability of landing on an unshaded section is the number of unshaded sections (3) divided by the total number of sections (6):
P(unshaded) = 3/6 = 1/2
Therefore, the probability of landing on an unshaded section in the sample space is 1/2.
So the answer is D: 1/2.
Which set of events is dependent?(1 point)
Responses
rolling a number cube and flipping a coin
rolling a number cube and flipping a coin
choosing a marble from a box, replacing it, then choosing another marble
choosing a marble from a box, replacing it, then choosing another marble
the event of getting two heads when flipping two fair coins
, 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
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 outcome of the first event affects the probability or outcome of the second event. If a ball is removed from the bag, it changes the probability or number of balls of each color left in the bag for the second draw.
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)
Responses
776
Start Fraction 7 over 76 End Fraction
21190
, Start Fraction 21 over 190 End Fraction
780
, Start Fraction 7 over 80 End Fraction
35
, Start Fraction 3 over 5 End Fraction
The probability of selecting an apple and a banana can be found by dividing the number of ways to select one apple and one banana by the total number of ways to select two fruits without replacement.
The number of ways to select one apple and one banana is:
7 (number of apples) x 5 (number of bananas) = 35
The total number of ways to select two fruits without replacement is:
20 (total number of fruits) x 19 (remaining fruits after first selection) = 380
Therefore, the probability of selecting an apple and a banana is:
35/380 = 7/76
So the answer is, Start Fraction 7 over 76 End Fraction.
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
75%
75%
12.5%
12.5%
0
0
62.5%
The probability of selecting an apple or a banana can be found by adding the individual probabilities of selecting an apple and selecting a banana.
The probability of selecting a banana is:
10 (number of bananas) / 20 (total number of fruits) = 1/2
The probability of selecting an apple is:
5 (number of apples) / 20 (total number of fruits) = 1/4
Therefore, the probability of selecting an apple or a banana is:
1/2 + 1/4 = 3/4
Multiplying by 100% to express the answer in percent form gives:
(3/4) x 100% = 75%
So the probability of selecting an apple or a banana is 75% in percent form.
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)
Responses
1325+1025
, Start Fraction 13 over 25 End Fraction plus Start Fraction 10 over 25 End Fraction
1325×1025
, Start Fraction 13 over 25 End Fraction times Start Fraction 10 over 25 End Fraction
325
Start Fraction 3 over 25 End Fraction
1325+1025−325
Start Fraction 13 over 25 End Fraction plus Start Fraction 10 over 25 End Fraction minus Start Fraction 3 over 25 End Fraction
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The probability of selecting a student who wants the option to keep studying online or is a boy can be found by adding the probability of selecting a student who wants the option to keep studying online to the probability of selecting a boy, then subtracting the probability of selecting a student who is both (since this would be counted twice in the first two probabilities).
Using the numbers in the table, we can calculate these probabilities as follows:
Probability of selecting a student who wants the option to keep studying online:
13/25
Probability of selecting a boy:
10/25 = 2/5
Probability of selecting a student who is both:
3/25
Therefore, the mathematical sentence that can be used to calculate the probability of selecting a student who wants the option to keep studying online or is a boy is:
13/25 + 2/5 - 3/25
Simplifying by finding a common denominator gives:
(13/25) + (10/25) - (3/25) = 20/25
This can be further simplified by dividing both the numerator and denominator by 5:
20/25 = 4/5
So the final answer is Start Fraction 4 over 5 End Fraction.
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?
A: Start Fraction 13 over 25 End Fraction plus Start Fraction 10 over 25 End Fraction
B: Start Fraction 13 over 25 End Fraction times Start Fraction 10 over 25 End Fraction
C: Start Fraction 3 over 25 End Fraction
D: Start Fraction 13 over 25 End Fraction plus Start Fraction 10 over 25 End Fraction minus Start Fraction 3 over 25 End Fraction
The correct mathematical sentence that can be used to calculate the probability of selecting a student who wants the option to keep studying online or is a boy is:
A: Start Fraction 13 over 25 End Fraction plus Start Fraction 10 over 25 End Fraction
This is because we want to find the probability of selecting either a student who wants the option to keep studying online or a boy, which means we need to add the probability of selecting a student who wants the option to keep studying online to the probability of selecting a boy.