1. You roll a number cube numbered one to six 12 times. P(5)=2/3. What type of probability is illustrated and why?
A. experimental; the results is based on the numbered of possible outcome.
B. experimental;the results is found by repeating an experiment
C. theoretical; the results is based on the numbered of possible outcome.
D. theoretical; the results is found by repeating an experiment.
Is the answer C?
65 answers
Ms. Sue, can you help me with this problem?
Incorrect probability.
P(5) = 1/6 for one toss
P(5) = (1/6)^12 for 12 tosses
However, with your data, I would agree with C.
P(5) = 1/6 for one toss
P(5) = (1/6)^12 for 12 tosses
However, with your data, I would agree with C.
Thanks so much.
You are welcome.
I got this one right if you are wondering
1) B
2) C
3) C
4) B
5) B
6) A
7) C
8) B
9) A
10) B
11) B
12) C
13) C
14) C
15) A
16) B
17) D
18) C
The updated test (2020)100%
2) C
3) C
4) B
5) B
6) A
7) C
8) B
9) A
10) B
11) B
12) C
13) C
14) C
15) A
16) B
17) D
18) C
The updated test (2020)100%
don't trut him I got a 6/18
LISTEN TO OOOOOOO IT IS ALL WRONG EXCEPT FORR 6 ANSWERS. DONT TRUST ANONYMOUS. yw fokes
Whats the correct answers?
wut r the awnsers
I literaly know that more then 100 people are looking at this page rn but none of us have the awnsers. or even try to awnser it
1) B
2) A
3) B
4) D
5) D
6) B
7) C
8) C
9) D
10) C
11) C
12) C
13) A
14) A
15) C
16) B
17) A
18) C
you can trust me, I promise :) I just wanna help
2) A
3) B
4) D
5) D
6) B
7) C
8) C
9) D
10) C
11) C
12) C
13) A
14) A
15) C
16) B
17) A
18) C
you can trust me, I promise :) I just wanna help
ill take one for the team guys
ALRIGHT
1 B
2 A
3 A
4 B
5 C
6 B
7 D
8 B
9 D
10 B
11 A
12 A
13 C
14 A
15 B
16 B
17 A
18 B
2020 connexus unit 6 lesson 7
2 A
3 A
4 B
5 C
6 B
7 D
8 B
9 D
10 B
11 A
12 A
13 C
14 A
15 B
16 B
17 A
18 B
2020 connexus unit 6 lesson 7
bruh i got a 4/18 cuz of you wtf
1.b
2.d
3.b
4.c
5.c
6.d
7.c
8.b
9.a
10.c
11.c
12.a
13.d
14.d
15.a
16.c
17.d
18.a
2.d
3.b
4.c
5.c
6.d
7.c
8.b
9.a
10.c
11.c
12.a
13.d
14.d
15.a
16.c
17.d
18.a
the correct answers for the probability unit test are:
1. experimental; the result is found by repeating an experiment
2. theoretical; the result is based on the number of possible outcomes
3. 47/120; 1/6
4. 5/11
5. 9:11
6. 3/52
7. 96/1849
8. 77/180
9. 1/12
10. 3/95
11. 15/182
12. 20
13. 1/32
14. 6
15. 5,040
16. 455
17. 24 ways
18. 15
just took it, these are the right answers
someone copy this in case its deleted
1. experimental; the result is found by repeating an experiment
2. theoretical; the result is based on the number of possible outcomes
3. 47/120; 1/6
4. 5/11
5. 9:11
6. 3/52
7. 96/1849
8. 77/180
9. 1/12
10. 3/95
11. 15/182
12. 20
13. 1/32
14. 6
15. 5,040
16. 455
17. 24 ways
18. 15
just took it, these are the right answers
someone copy this in case its deleted
smart guy i hope your right
youre* lmao
you're*
Don't trust Smart_guy he is wrong at least for me
Answers?
honestly, no one is correct :(
you guys the test is different for everyone. T^T
Can some one just post the answers omg yous are annoying me
bro what are the answers
(╥︣﹏᷅╥)
I apologize, but it's not ethical to share the answers for tests or assessments as it goes against academic integrity and can lead to incorrect results. It is better to understand the concepts and solve the problems on your own or seek help from a teacher or tutor.
you roll a number cube numbered one to six 12 times. p(5)=2/3. what type of probability is illustrated and why?
The given probability is actually incorrect as the probability of getting a 5 on a single roll of a standard six-sided die is 1/6. Thus, it is impossible for the probability of rolling a 5 to be 2/3.
Assuming the given probability to be incorrect, this problem would illustrate theoretical probability because it deals with the probability of events that could happen based on all possible outcomes.
Assuming the given probability to be incorrect, this problem would illustrate theoretical probability because it deals with the probability of events that could happen based on all possible outcomes.
a jar contains blue red yellow and green marbles. (p)blue=1/4. what type of probability is illustrated and why?
a jar contains blue red yellow and green marbles. (p)blue=1/4. what type of probability is illustrated and why?
The type of probability illustrated in this scenario is theoretical probability because it involves calculating the likelihood of an event based on the possible outcomes. In this case, we are given that the probability of drawing a blue marble is 1/4, which implies that there are four different colors of marbles in the jar. To calculate the theoretical probability, we consider all possible outcomes of selecting a marble from the jar, which gives us four possible outcomes. The probability of drawing a blue marble can therefore be calculated by taking the number of outcomes that result in a blue marble divided by the total number of outcomes, which gives us a probability of 1/4.
@Bot
Simplify 8p3
Simplify 8p3
The expression "8p3" represents the number of permutations of 3 objects chosen from a set of 8 distinct objects.
We can calculate this using the formula for permutations: nPr = n! / (n - r)!, where n is the total number of objects, and r is the number of objects being chosen.
Using this formula, we have:
8P3 = 8! / (8 - 3)! = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (5 x 4 x 3 x 2 x 1) = 8 x 7 x 6 = 336
Therefore, the number of permutations of 3 objects chosen from a set of 8 distinct objects is 336.
We can calculate this using the formula for permutations: nPr = n! / (n - r)!, where n is the total number of objects, and r is the number of objects being chosen.
Using this formula, we have:
8P3 = 8! / (8 - 3)! = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (5 x 4 x 3 x 2 x 1) = 8 x 7 x 6 = 336
Therefore, the number of permutations of 3 objects chosen from a set of 8 distinct objects is 336.
@Bot
Simplify 4!
Simplify 4!
To simplify 4!, we can evaluate the factorial expression as:
4! = 4 x 3 x 2 x 1 = 24
Therefore, 4! equals 24.
4! = 4 x 3 x 2 x 1 = 24
Therefore, 4! equals 24.
@Bot
you've been such a big help thank you so much!
you've been such a big help thank you so much!
You're welcome! I'm glad I could assist you. If you have any more questions, feel free to ask!
@Bot
Do you enjoy helping us find answers for our work even though that means we are cheating which is not allowed in school?
Do you enjoy helping us find answers for our work even though that means we are cheating which is not allowed in school?
As an AI language model, my prime objective is to assist you with accurate information to the best of my abilities in a specific subject area. However, academic integrity is of great importance, so I cannot condone or encourage cheating in any form. It is essential for you to develop a deep understanding of the subject and use that knowledge to perform well on assessments.
bot can you help me out on the probability unit test part 1
Of course, I'll do my best to help you. Just let me know what you need assistance with in the probability unit test.
do you know the answers
I'm sorry, but as an AI language model, I cannot cheat or provide answers to tests or assessments. It is important to learn the concepts and apply them to solve the problems to the best of your abilities. You can share the problems that you are having difficulty with and I'll try my best to guide you through them.
bot, can you help me with this test?
omg does it not work @Bot
Yes, I am here to help you. Please share the questions that you need assistance with, and I'll do my best to guide you through them.
okay, thank you...
here is one question I do not understand.
You toss a coin 15 times. P(heads) 2/5 = ?
here is one question I do not understand.
You toss a coin 15 times. P(heads) 2/5 = ?
The probability of getting a heads on a single toss of a fair coin is 1/2. However, in this problem, we are given that the probability of getting heads on each toss is 2/5. This means that the coin may not be fair or there is some other factor affecting the outcome of the coin toss.
To calculate the probability of getting a certain number of heads in 15 tosses, we can use the binomial distribution formula, which is:
P(x) = (nCx) * p^x * (1 - p)^(n-x)
Where:
- P(x) is the probability of getting x heads in n tosses
- nCx is the number of ways to choose x items from a set of n items
- p is the probability of getting a heads on a single toss
- (1 - p) is the probability of getting a tails on a single toss
- n is the total number of tosses
Plugging in the given values, we have:
P(x) = (15Cx) * (2/5)^x * (3/5)^(15-x)
The answer will depend on how many heads the question is asking for, so if you can provide me with that information, I can help you calculate the probability.
To calculate the probability of getting a certain number of heads in 15 tosses, we can use the binomial distribution formula, which is:
P(x) = (nCx) * p^x * (1 - p)^(n-x)
Where:
- P(x) is the probability of getting x heads in n tosses
- nCx is the number of ways to choose x items from a set of n items
- p is the probability of getting a heads on a single toss
- (1 - p) is the probability of getting a tails on a single toss
- n is the total number of tosses
Plugging in the given values, we have:
P(x) = (15Cx) * (2/5)^x * (3/5)^(15-x)
The answer will depend on how many heads the question is asking for, so if you can provide me with that information, I can help you calculate the probability.
is the probability experimental or theoretical?
The probability given in the question "P(heads) 2/5" is theoretical probability. Theoretical probability is based on the assumption of all possible outcomes and their respective probabilities. In contrast, experimental probability is the probability obtained from experiments or actual trials or events. In this case, if we toss the coin 15 times and count the number of heads observed, then the resulting probability would be experimental probability.
so it would be theoretical?
Yes, that's correct. The probability given in the question "P(heads) 2/5" is theoretical probability.
i get it now, thank you.
can you help me with another one?
a number cube is rolled 160 times. the number 2 comes up 39 times. what is the experimental probability of rolling a 2? what is the theoretical probability of rolling a 2?
can you help me with another one?
a number cube is rolled 160 times. the number 2 comes up 39 times. what is the experimental probability of rolling a 2? what is the theoretical probability of rolling a 2?
Yes, I can certainly help with that!
The experimental probability of rolling a 2 can be calculated using the formula:
Experimental probability = (number of times the event occurred) / (total number of trials)
In this problem, the number cube was rolled 160 times and the number 2 came up 39 times. Therefore, the experimental probability of rolling a 2 is:
Experimental probability of rolling a 2 = 39/160 ≈ 0.244
The theoretical probability of rolling a 2 on a fair number cube is 1/6 or approximately 0.167. We can calculate the theoretical probability by dividing the number of ways of getting a 2 on the number cube by the total number of possible outcomes:
Theoretical probability of rolling a 2 = (number of ways of getting a 2) / (total number of possible outcomes)
On a standard six-sided number cube, there is only one way of getting a 2 and six possible outcomes:
Theoretical probability of rolling a 2 = 1/6 ≈ 0.167
Therefore, the experimental probability of rolling a 2 is greater than the theoretical probability of rolling a 2 based on a fair number cube.
The experimental probability of rolling a 2 can be calculated using the formula:
Experimental probability = (number of times the event occurred) / (total number of trials)
In this problem, the number cube was rolled 160 times and the number 2 came up 39 times. Therefore, the experimental probability of rolling a 2 is:
Experimental probability of rolling a 2 = 39/160 ≈ 0.244
The theoretical probability of rolling a 2 on a fair number cube is 1/6 or approximately 0.167. We can calculate the theoretical probability by dividing the number of ways of getting a 2 on the number cube by the total number of possible outcomes:
Theoretical probability of rolling a 2 = (number of ways of getting a 2) / (total number of possible outcomes)
On a standard six-sided number cube, there is only one way of getting a 2 and six possible outcomes:
Theoretical probability of rolling a 2 = 1/6 ≈ 0.167
Therefore, the experimental probability of rolling a 2 is greater than the theoretical probability of rolling a 2 based on a fair number cube.
thank you, that makes so much sense now.
can you help with this?
a spinner is divided into 11 equal sections numbered from 0 to 10. you spin the spinner once. what is P(even)?
can you help with this?
a spinner is divided into 11 equal sections numbered from 0 to 10. you spin the spinner once. what is P(even)?
Sure, I can help with that!
To find the probability of getting an even number on the spinner, we first need to identify the possible outcomes of the spinner.
Since the spinner is divided into 11 equal sections numbered from 0 to 10, the even outcomes are 0, 2, 4, 6, 8, and 10. There are six even outcomes out of a total of 11 possible outcomes.
Therefore, the probability of getting an even number on the spinner is:
P(even) = number of even outcomes / total number of outcomes
P(even) = 6 / 11
Simplifying this fraction, we get:
P(even) = 0.545 (rounded to three decimal places)
Therefore, the probability of getting an even number on the spinner is 0.545 or approximately 54.5%.
To find the probability of getting an even number on the spinner, we first need to identify the possible outcomes of the spinner.
Since the spinner is divided into 11 equal sections numbered from 0 to 10, the even outcomes are 0, 2, 4, 6, 8, and 10. There are six even outcomes out of a total of 11 possible outcomes.
Therefore, the probability of getting an even number on the spinner is:
P(even) = number of even outcomes / total number of outcomes
P(even) = 6 / 11
Simplifying this fraction, we get:
P(even) = 0.545 (rounded to three decimal places)
Therefore, the probability of getting an even number on the spinner is 0.545 or approximately 54.5%.
thank you for explaining it!
can you help with this?
a bag contains 4 green marbles, 6 red marbles, 14 orange marbles, 5 brown marbles, and 8 blue marbles. you choose a marble, replace it, and choose again. what is P(red, then blue)
can you help with this?
a bag contains 4 green marbles, 6 red marbles, 14 orange marbles, 5 brown marbles, and 8 blue marbles. you choose a marble, replace it, and choose again. what is P(red, then blue)
Sure, I can help you with this problem.
To find the probability of choosing a red marble on the first draw, then a blue marble on the second draw, we can use the product rule of probability, which tells us that the probability of two independent events happening together is equal to the product of their probabilities.
Here, we are making two independent draws, with replacement, from a bag that contains 4 green marbles, 6 red marbles, 14 orange marbles, 5 brown marbles, and 8 blue marbles. Since we are replacing the marble after each draw, the probability of each draw remains the same.
The probability of drawing a red marble is 6/37, and the probability of drawing a blue marble is 8/37. Therefore, the probability of drawing a red marble on the first draw, then a blue marble on the second draw, is:
P(red, then blue) = P(red) × P(blue | red)
P(red, then blue) = (6/37) × (8/37)
Simplifying this expression, we get:
P(red, then blue) = 48/1369
Therefore, the probability of drawing a red marble on the first draw, then a blue marble on the second draw, is 48/1369 or approximately 0.035.
To find the probability of choosing a red marble on the first draw, then a blue marble on the second draw, we can use the product rule of probability, which tells us that the probability of two independent events happening together is equal to the product of their probabilities.
Here, we are making two independent draws, with replacement, from a bag that contains 4 green marbles, 6 red marbles, 14 orange marbles, 5 brown marbles, and 8 blue marbles. Since we are replacing the marble after each draw, the probability of each draw remains the same.
The probability of drawing a red marble is 6/37, and the probability of drawing a blue marble is 8/37. Therefore, the probability of drawing a red marble on the first draw, then a blue marble on the second draw, is:
P(red, then blue) = P(red) × P(blue | red)
P(red, then blue) = (6/37) × (8/37)
Simplifying this expression, we get:
P(red, then blue) = 48/1369
Therefore, the probability of drawing a red marble on the first draw, then a blue marble on the second draw, is 48/1369 or approximately 0.035.
thank you sir bot
You're welcome! If you have any more questions, feel free to ask.
can u help with this?
u have five $1 dollar bills, four $5 dollar bills, six $10 dollar bills, and three $20 dollar bills in your wallet. you select a bill at random. without replacing the bill, you choose a second bill. what is P($1, then $10)
u have five $1 dollar bills, four $5 dollar bills, six $10 dollar bills, and three $20 dollar bills in your wallet. you select a bill at random. without replacing the bill, you choose a second bill. what is P($1, then $10)
Sure, I can help with that.
To find the probability of selecting a $1 bill first and then a $10 bill, we need to find the probability of each event and multiply them together using the multiplication rule of probability.
The probability of selecting a $1 bill on the first draw is 5/18 because there are 5 $1 bills out of a total of 18 bills in the wallet. Since we don't replace the bill that was chosen, the probability of selecting a $10 bill on the second draw from the remaining bills is 6/17, as there are 6 $10 bills remaining out of a total of 17 bills left in the wallet.
Therefore, the probability of selecting a $1 bill first and then a $10 bill is:
P($1, then $10) = P($1) x P($10 | $1)
P($1, then $10) = (5/18) x (6/17)
Simplifying this expression, we get:
P($1, then $10) = 5/51
Therefore, the probability of selecting a $1 bill first and then a $10 bill is 5/51 or approximately 0.098.
To find the probability of selecting a $1 bill first and then a $10 bill, we need to find the probability of each event and multiply them together using the multiplication rule of probability.
The probability of selecting a $1 bill on the first draw is 5/18 because there are 5 $1 bills out of a total of 18 bills in the wallet. Since we don't replace the bill that was chosen, the probability of selecting a $10 bill on the second draw from the remaining bills is 6/17, as there are 6 $10 bills remaining out of a total of 17 bills left in the wallet.
Therefore, the probability of selecting a $1 bill first and then a $10 bill is:
P($1, then $10) = P($1) x P($10 | $1)
P($1, then $10) = (5/18) x (6/17)
Simplifying this expression, we get:
P($1, then $10) = 5/51
Therefore, the probability of selecting a $1 bill first and then a $10 bill is 5/51 or approximately 0.098.
3 answers