I’m fairly sure it’s increasing, I just can’t tell by how much

Answer:

See explanation below

Step-by-step explanation:

Here a coin was tossed three times.

Let H = head &  T = tail

Find the following:

a) The sample space:

Since a coin is tossed thrice, all possible outcome would be:

S = { HHH, HHT, HTH, HTT, TTT, TTH, THH, THT}

b) i) A = Exactly 2 tails: Here exactly 2 tails were recorded.

A = {HTT, TTH, THT}

ii) B = at least two tails: Here 2 or more tails were recorded.

B = {HTT, TTT, TTH, THT}

iii) C = the last two tosses are heads:

C = { HHH, THH}

c) List the elements of the following events:

i) A. This means all outcomes in A

= {HTT, TTH, THT}

ii) A∪B. A union B, means all possible outcomes present in A or B or in both

= {HTT, TTH, THT, TTT}

iii) A∩B. This means all possible outcomes of A that are present in B.

= {HTT, TTH, THT}

iv) A∩C. All outcomes A that are present in B

= {∅}

The sample space of tossing a coin three times consists of eight possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT. Events A, B, and C can be determined by listing the appropriate outcomes. The intersection and union of events A and B can also be determined.

(a) The sample space, Ω, of tossing a coin three times can be determined by listing all the possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT.

(b) i. A = {HHT, HTH, THH}

ii. B = {TTT, TTH, THT, HTT, HHT, HTH, THH}

iii. C = {HTH, TTH}

(c) i. A = {HHT, HTH, THH}

ii. A∪B = {HHT, HTH, THH, TTT, TTH, THT, HTT, HHT}

iii. A∩B = {HHT, HTH, THH}

iv. A∩C = {HHT, HTH}

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Answer:

The answer is "20".

Step-by-step explanation:

It is also known as the group of the study, that targets the population, which helps to find the survey, which is the sampled population. It is measured by an ideal world, which will be the same, and they're always unique.  

• Its sampling distribution of the "x bar" should also be naturally independent of the random sample, that is usually distributed.
• We consider the population as endless if the sampling size is at least 20 times greater than the sample size.

Answer:

The answer is 2

Step-by-step explanation:

Use a calculator.

Answer:

Step-by-step explanation:

https://tex.z-dn.net/?f=%5Csigma%20_%7Bmean%7D%3D70%20MPa%3D%5Cfrac%7B%5Csigma%20_%7Bmax%7D%2B%5Csigma%20_%7Bmin%7D%7D%7B2%7D

Answer:

The probability that he or she is high-risk is 0.50

Step-by-step explanation:

P(Low risk) = 40% = 0.40

P( Moderate risk) = 40% = 0.40

P(High risk) = 20% = 0.20

P(At - fault accident | Low risk) = 0% = 0

P(At-fault accident | Moderate risk) = 10% = 0.10  

P(At-fault accident | High risk) = 20% = 0.20

If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk. Hence, We need to  calculate P( High risk | at-fault accident) = ?

Using Bayes' conditional probability theorem

P( High risk | at-fault accident) = ( P( High risk) * P(At-fault accident | High risk) ) /  { P( Low risk) * P(At-fault accident | Low risk) +P( Moderate risk) * P(At-fault accident | Moderate risk) +  P( High risk) * P(At-fault accident | High risk) }

P( High risk | at-fault accident)= (0.20 * 0.20) / ( 0.40 * 0 + 0.40 * 0.10 + 0.20 * 0.20 )

P( High risk | at-fault accident) = 0.04 / 0 + 0.04 + 0.04

P( High risk | at-fault accident) = 0.04 / 0.08

P( High risk | at-fault accident) = 0.50.

Final answer:

The probability that a driver is high-risk given that they had an at-fault accident can be found using Bayes' theorem. Given the probabilities provided in the question, the probability is approximately 0.3333 or 33.33%.

Explanation:

To find the probability that a driver is high-risk given that they had an at-fault accident, we can use Bayes' theorem. Let's define the events:

1. A: Driver is high-risk
2. B: Driver has an at-fault accident

We are given the following probabilities:

1. P(A) = 0.20 (probability of a driver being high-risk)
2. P(B|A) = 0.20 (probability of an at-fault accident given that they are high-risk)
3. P(~A) = 0.80 (probability of a driver not being high-risk)
4. P(B|~A) = 0.10 (probability of an at-fault accident given that they are not high-risk)

Using Bayes' theorem, the probability of a driver being high-risk given that they had an at-fault accident is:

P(A|B) = (P(A) * P(B|A)) / (P(A) * P(B|A) + P(~A) * P(B|~A))

Substituting the given probabilities:

P(A|B) = (0.20 * 0.20) / (0.20 * 0.20 + 0.80 * 0.10) = 0.04 / (0.04 + 0.08) = 0.04 / 0.12 = 0.3333.

Therefore, the probability that a driver is high-risk given that they had an at-fault accident in the next year is approximately 0.3333 or 33.33%.

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hi

Seven Balls , one of each color.  Then 1/7 possibility to choose purple