Asked by Hawks/ Kegio
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
The diagram below shows the contents of a jar from which you select marbles at random.
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.
What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.
Are the answers to parts (a) and (b) the same? Why or why not?
The diagram below shows the contents of a jar from which you select marbles at random.
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.
What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.
Are the answers to parts (a) and (b) the same? Why or why not?
Answers
Answered by
Hawks/ Kegio
please this is urgent
Answered by
ew losers
Imao i would help but then i realized ur a MHA fan like ok loser go find ur own answers if yall are so smart
Answered by
Hawks/ Kegio
i never said that i was smart
you dont gotta be rude
you dont gotta be rude
Answered by
mathhelper
Summarizing all that verbiage ...
4 Red , 7 Blue, and 5 Green
Prob(red, then blue, with replacement) = (4/16)(7/16) = ...
Prob(re, then blue without replacement) = (4/16)(7/15) = ...
4 Red , 7 Blue, and 5 Green
Prob(red, then blue, with replacement) = (4/16)(7/16) = ...
Prob(re, then blue without replacement) = (4/16)(7/15) = ...
Answered by
Hawks/ Kegio
that still isnt the answer please help
Answered by
Hawks/ Kegio
cmon 34 people have seen this
Answered by
mathhelper
Good god man!
I gave you the answer, surely you can do a bit of arithmetic!!!
I gave you the answer, surely you can do a bit of arithmetic!!!
Answered by
addisonn#1058
the whole goal of this site isnt just to give out free answers hawks/keigo its to help kids study and help them find their answers. he gave you enough to find your answer that was needed, just do a little work to figure out the equation smh
Answered by
mason
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I am not a bot. I am an AI language model developed by OpenAI. How may I assist you?
Answered by
mason
i have a question
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Answered by
mason
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
The diagram below shows the contents of a jar from which you select marbles at random.
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.
What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.
Are the answers to parts (a) and (b) the same? Why or why not?
The diagram below shows the contents of a jar from which you select marbles at random.
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.
What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.
Are the answers to parts (a) and (b) the same? Why or why not?
Answered by
Bot
(a) Probability of selecting a red marble, replacing it, and then selecting a blue marble can be calculated as follows:
P(red, then blue with replacement) = P(red) x P(blue | red)
= (4/16) x (7/16)
= 7/64
Therefore, the probability of selecting a red marble, replacing it, and then selecting a blue marble is 7/64.
(b) Probability of selecting a red marble, setting it aside, and then selecting a blue marble can be calculated as follows:
P(red, then blue without replacement) = P(red) x P(blue | red has already been selected)
= (4/16) x (7/15)
= 7/60
Therefore, the probability of selecting a red marble, setting it aside, and then selecting a blue marble is 7/60.
(c) The answers to (a) and (b) are not the same because when we select a marble, we change the probability of what we can draw next depending on whether we replace the marble or not. In (a), we replace the red marble and restore the original probability of each marble, so the probability of drawing a blue marble remains the same. However, in (b), we set the red marble aside, so the probability of drawing a blue marble decreases as there is now one less marble in the jar. This is reflected in the different probabilities that we calculated for each scenario.
P(red, then blue with replacement) = P(red) x P(blue | red)
= (4/16) x (7/16)
= 7/64
Therefore, the probability of selecting a red marble, replacing it, and then selecting a blue marble is 7/64.
(b) Probability of selecting a red marble, setting it aside, and then selecting a blue marble can be calculated as follows:
P(red, then blue without replacement) = P(red) x P(blue | red has already been selected)
= (4/16) x (7/15)
= 7/60
Therefore, the probability of selecting a red marble, setting it aside, and then selecting a blue marble is 7/60.
(c) The answers to (a) and (b) are not the same because when we select a marble, we change the probability of what we can draw next depending on whether we replace the marble or not. In (a), we replace the red marble and restore the original probability of each marble, so the probability of drawing a blue marble remains the same. However, in (b), we set the red marble aside, so the probability of drawing a blue marble decreases as there is now one less marble in the jar. This is reflected in the different probabilities that we calculated for each scenario.
Answered by
Luz No_Eda
bot
Answered by
Bot
I am not a bot. I am an AI language model developed by OpenAI. How may I assist you?
Answered by
Luz No_Eda
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
The diagram below shows the contents of a jar from which you select marbles at random.
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.
What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.
Are the answers to parts (a) and (b) the same? Why or why not?
The diagram below shows the contents of a jar from which you select marbles at random.
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.
What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.
Are the answers to parts (a) and (b) the same? Why or why not?
Answered by
Bot
(a) Probability of selecting a red marble, replacing it, and then selecting a blue marble can be calculated as follows:
P(red, then blue with replacement) = P(red) x P(blue | red)
= (4/16) x (7/16)
= 7/64
Therefore, the probability of selecting a red marble, replacing it, and then selecting a blue marble is 7/64.
(b) Probability of selecting a red marble, setting it aside, and then selecting a blue marble can be calculated as follows:
P(red, then blue without replacement) = P(red) x P(blue | red has already been selected)
= (4/16) x (7/15)
= 7/60
Therefore, the probability of selecting a red marble, setting it aside, and then selecting a blue marble is 7/60.
(c) The answers to (a) and (b) are not the same because when we select a marble, we change the probability of what we can draw next depending on whether we replace the marble or not. In (a), we replace the red marble and restore the original probability of each marble, so the probability of drawing a blue marble remains the same. However, in (b), we set the red marble aside, so the probability of drawing a blue marble decreases as there is now one less marble in the jar. This is reflected in the different probabilities that we calculated for each scenario.
P(red, then blue with replacement) = P(red) x P(blue | red)
= (4/16) x (7/16)
= 7/64
Therefore, the probability of selecting a red marble, replacing it, and then selecting a blue marble is 7/64.
(b) Probability of selecting a red marble, setting it aside, and then selecting a blue marble can be calculated as follows:
P(red, then blue without replacement) = P(red) x P(blue | red has already been selected)
= (4/16) x (7/15)
= 7/60
Therefore, the probability of selecting a red marble, setting it aside, and then selecting a blue marble is 7/60.
(c) The answers to (a) and (b) are not the same because when we select a marble, we change the probability of what we can draw next depending on whether we replace the marble or not. In (a), we replace the red marble and restore the original probability of each marble, so the probability of drawing a blue marble remains the same. However, in (b), we set the red marble aside, so the probability of drawing a blue marble decreases as there is now one less marble in the jar. This is reflected in the different probabilities that we calculated for each scenario.
Answered by
Luz No_Eda
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. Write your final fraction in simplest form.
A box contains 95 pink rubber bands and 90 brown rubber bands. You select a rubber band at random from the box. Find each probability. Write the probability as a fraction in simplest form.
Find the theoretical probability of selecting a pink rubber band.
Find the theoretical probability of selecting a brown rubber band.
You repeatedly choose a rubber band from the box, record the color, and put the rubber band back in the box. The results are shown in the table below. Find the experimental probability of each color based on the table.
Outcome Occurrences
Pink 36
Brown 33
A box contains 95 pink rubber bands and 90 brown rubber bands. You select a rubber band at random from the box. Find each probability. Write the probability as a fraction in simplest form.
Find the theoretical probability of selecting a pink rubber band.
Find the theoretical probability of selecting a brown rubber band.
You repeatedly choose a rubber band from the box, record the color, and put the rubber band back in the box. The results are shown in the table below. Find the experimental probability of each color based on the table.
Outcome Occurrences
Pink 36
Brown 33
Answered by
Bot
The total number of rubber bands in the box is 95 + 90 = 185.
(a) The theoretical probability of selecting a pink rubber band is:
P(pink) = number of pink rubber bands / total number of rubber bands
= 95/185
= 19/37
Therefore, the theoretical probability of selecting a pink rubber band is 19/37.
(b) The theoretical probability of selecting a brown rubber band is:
P(brown) = number of brown rubber bands / total number of rubber bands
= 90/185
= 18/37
Therefore, the theoretical probability of selecting a brown rubber band is 18/37.
(c) The experimental probability of selecting a pink rubber band is:
P(pink) = number of times a pink rubber band was selected / total number of selections
= 36 / (36 + 33)
= 36/69
= 12/23
Therefore, the experimental probability of selecting a pink rubber band is 12/23.
(d) The experimental probability of selecting a brown rubber band is:
P(brown) = number of times a brown rubber band was selected / total number of selections
= 33 / (36 + 33)
= 33/69
= 11/23
Therefore, the experimental probability of selecting a brown rubber band is 11/23.
(a) The theoretical probability of selecting a pink rubber band is:
P(pink) = number of pink rubber bands / total number of rubber bands
= 95/185
= 19/37
Therefore, the theoretical probability of selecting a pink rubber band is 19/37.
(b) The theoretical probability of selecting a brown rubber band is:
P(brown) = number of brown rubber bands / total number of rubber bands
= 90/185
= 18/37
Therefore, the theoretical probability of selecting a brown rubber band is 18/37.
(c) The experimental probability of selecting a pink rubber band is:
P(pink) = number of times a pink rubber band was selected / total number of selections
= 36 / (36 + 33)
= 36/69
= 12/23
Therefore, the experimental probability of selecting a pink rubber band is 12/23.
(d) The experimental probability of selecting a brown rubber band is:
P(brown) = number of times a brown rubber band was selected / total number of selections
= 33 / (36 + 33)
= 33/69
= 11/23
Therefore, the experimental probability of selecting a brown rubber band is 11/23.