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The probability that at most 8 of them take the bus to school is 0.925, written in percentage form this is 92.5%
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How to find the probability?</h3>
We know that roughly 75% of the students take the bus, then, if we select a student at random.
- There is a probability of 0.75 that the student takes the bus.
- There is a probability of 0.25 that the student does not take the bus.
The probability that at most 8 out of 9 students take the bus, is equal to one minus the probability of the 9 taking the bus, which is:
p = (0.75)^9 = 0.075
Then we have:
P = 1 - 0.075 = 0.925
The probability that at most 8 of them take the bus to school is 0.925, written in percentage form this is 92.5%
If you want to learn more about probability, you can read:
brainly.com/question/251701
Answer:
51x
Step-by-step explanation:
3x^3 - 7x + 3x^3 + 4x
27x - 7x + 27x + 4x
20x + 31x
51x
Answer:
this is false. that would be the formula to find the volume
Step-by-step explanation:
Answer:
Correct option: (a) 0.1452
Step-by-step explanation:
The new test designed for detecting TB is being analysed.
Denote the events as follows:
<em>D</em> = a person has the disease
<em>X</em> = the test is positive.
The information provided is:

Compute the probability that a person does not have the disease as follows:

The probability of a person not having the disease is 0.12.
Compute the probability that a randomly selected person is tested negative but does have the disease as follows:
![P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%29%3DP%28X%5E%7Bc%7D%7CD%29P%28D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%20P%28D%29%5C%5C%3D%5B1-0.97%5D%5Ctimes%200.88%5C%5C%3D0.03%5Ctimes%200.88%5C%5C%3D0.0264)
Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:
![P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%5E%7Bc%7D%29%3DP%28X%5E%7Bc%7D%7CD%5E%7Bc%7D%29P%28D%5E%7Bc%7D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%7B1-%20P%28D%29%5D%5C%5C%3D0.99%5Ctimes%200.12%5C%5C%3D0.1188)
Compute the probability that a randomly selected person is tested negative as follows:


Thus, the probability of the test indicating that the person does not have the disease is 0.1452.