B. Negative exponent flip from denominator to numerator and numerator to denominator.
B is the second option.
Total number of 10 grader students = 260 students.
Total number of students who rides the bus and they are of 10th grader = 150 students.
Probability (a student rides the bus given that they are a 10th grader) =![\frac{Total \ number \ of \ students \ rides \ the \ bus \ and \ 10th \ grader}{Total \ number \ of \ 10 \ grader \ students}](https://tex.z-dn.net/?f=%5Cfrac%7BTotal%20%5C%20number%20%5C%20of%20%5C%20students%20%5C%20rides%20%5C%20the%20%5C%20bus%20%5C%20and%20%5C%2010th%20%5C%20grader%7D%7BTotal%20%5C%20number%20%5C%20of%20%5C%2010%20%5C%20grader%20%5C%20students%7D)
Therefore,
P(E) = ![\frac{150}{260}](https://tex.z-dn.net/?f=%5Cfrac%7B150%7D%7B260%7D)
Dividing top and bottom by 10, we get
P(E) = 15/26.
In decimals 0.577.
Therrefore, correct option is C) 150/260 = 15/26 = .577.
Answer:
Correct option: A. P=0.057
Step-by-step explanation:
<u>Binomial Distribution</u>
The binomial distribution fits the case of n independent events each one with a probability of success equal to p where k successes have been achieved.
The PMF (Probability Mass Function) of the binomial distribution is
![\displaystyle P(k,n,p)=\binom{n}{k}p^kq^{n-k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20P%28k%2Cn%2Cp%29%3D%5Cbinom%7Bn%7D%7Bk%7Dp%5Ekq%5E%7Bn-k%7D)
Where ![q = 1-p](https://tex.z-dn.net/?f=q%20%3D%201-p)
The individual probability that an electronic device does not function properly is p=0.1. We know n=10 devices have been bought and we want to compute the probability that 7 devices function properly. Please notice that this is not the value of k since p is the probability of failure. The correct value of k=10-7=3. The required probability is
![\displaystyle P(3,10,0.1)=\binom{10}{3}0.1^30.9^{10-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20P%283%2C10%2C0.1%29%3D%5Cbinom%7B10%7D%7B3%7D0.1%5E30.9%5E%7B10-3%7D)
![\boxed{P=0.057}](https://tex.z-dn.net/?f=%5Cboxed%7BP%3D0.057%7D)