Answer:
a.![P(E_1/A)=0.0789](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D0.0789)
b.
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c.![P(E_3/A)=0.526](https://tex.z-dn.net/?f=P%28E_3%2FA%29%3D0.526)
Step-by-step explanation:
Let
are the events that denotes the good drive, medium drive and poor risk driver.
![P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20](https://tex.z-dn.net/?f=P%28E_1%29%3D0.30%2CP%28E_2%29%3D0.50%2CP%28E_3%29%3D0.20)
Let A be the event that denotes an accident.
![P(A/E_1)=0.01](https://tex.z-dn.net/?f=P%28A%2FE_1%29%3D0.01)
![P(A/E_2=0.03](https://tex.z-dn.net/?f=P%28A%2FE_2%3D0.03)
![P(A/E_3)=0.10](https://tex.z-dn.net/?f=P%28A%2FE_3%29%3D0.10)
The company sells Mr. Brophyan insurance policy and he has an accident.
a.We have to find the probability Mr.Brophy is a good driver
Bayes theorem,![P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}](https://tex.z-dn.net/?f=P%28E_i%2FA%29%3D%5Cfrac%7BP%28A%2FE_i%29%5Ccdot%20P%28E_1%29%7D%7B%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7DP%28A%2FE_i%29%5Ccdot%20P%28E_i%29%7D)
We have to find ![P(E_1/A)](https://tex.z-dn.net/?f=P%28E_1%2FA%29)
Using the Bayes theorem
![P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D%5Cfrac%7BP%28A%2FE_1%29%5Ccdot%20P%28E_1%29%7D%7BP%28E_1%29%5Ccdot%20P%28A%2FE_1%29%2BP%28E_2%29P%28A%2FE_2%29%2BP%28E_3%29P%28A%2FE_3%29%7D)
Substitute the values then we get
![P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D%5Cfrac%7B0.30%5Ctimes%200.01%7D%7B0.01%5Ctimes%200.30%2B0.50%5Ctimes%200.03%2B0.20%5Ctimes%200.10%7D)
![P(E_1/A)=0.0789](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D0.0789)
b.We have to find the probability Mr.Brophy is a medium driver
![P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395](https://tex.z-dn.net/?f=P%28E_2%2FA%29%3D%5Cfrac%7B0.03%5Ctimes%200.50%7D%7B0.038%7D%3D0.395)
c.We have to find the probability Mr.Brophy is a poor driver
![P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526](https://tex.z-dn.net/?f=P%28E_3%2FA%29%3D%5Cfrac%7B0.20%5Ctimes%200.10%7D%7B0.038%7D%3D0.526)
I think the answer might be x=4/8
Answer
given,
on first stop
number of car = 20 and number of trucks = 18
on second stop
number of car = 18 and number of trucks = 10
we need to calculate which rest stop has higher ratio of car to truck.
Rest Stop 1
ratio= r₁ =![\dfrac{cars}{trucks}](https://tex.z-dn.net/?f=%5Cdfrac%7Bcars%7D%7Btrucks%7D)
r₁ =![\dfrac{20}{18}](https://tex.z-dn.net/?f=%5Cdfrac%7B20%7D%7B18%7D)
r₁ =![\dfrac{10}{9}](https://tex.z-dn.net/?f=%5Cdfrac%7B10%7D%7B9%7D)
Rest Stop 2
ratio= r₂ =![\dfrac{cars}{trucks}](https://tex.z-dn.net/?f=%5Cdfrac%7Bcars%7D%7Btrucks%7D)
r₂ =![\dfrac{18}{10}](https://tex.z-dn.net/?f=%5Cdfrac%7B18%7D%7B10%7D)
r₂=![\dfrac{9}{5}](https://tex.z-dn.net/?f=%5Cdfrac%7B9%7D%7B5%7D)
hence, r₂ > r₁
rest stop 2 has more car to truck ratio than rest stop 1
(x+4) + ((x-15)+4)= 41
2x-7= 41
2x= 48
x= 24
Answer:
the sum is 25 not 24?
Step-by-step explanation: