Answer:
The probability that a randomly selected automobile manufactured by the company will have a defective microprocessor is 6.2 %.
Step-by-step explanation:
Given:
The probability of getting a microprocessor from firm A is, ![P(A)=45\%=0.45](https://tex.z-dn.net/?f=P%28A%29%3D45%5C%25%3D0.45)
The probability of getting a microprocessor from firm B is, ![P(B)=30\%=0.30](https://tex.z-dn.net/?f=P%28B%29%3D30%5C%25%3D0.30)
The probability of getting a microprocessor from firm C is, ![P(C)=25\%=0.25](https://tex.z-dn.net/?f=P%28C%29%3D25%5C%25%3D0.25)
Now, let event D be having a defective microprocessor at random.
So, as per the question,
Probability of producing a defective microprocessor from firm A is, ![P(D/A)=6\%=0.06](https://tex.z-dn.net/?f=P%28D%2FA%29%3D6%5C%25%3D0.06)
Probability of producing a defective microprocessor from firm B is, ![P(D/B)=7\%=0.07](https://tex.z-dn.net/?f=P%28D%2FB%29%3D7%5C%25%3D0.07)
Probability of producing a defective microprocessor from firm C is, ![P(D/C)=5.5\%=0.055](https://tex.z-dn.net/?f=P%28D%2FC%29%3D5.5%5C%25%3D0.055)
Now, probability of having a defective microprocessor when selected at random is given as:
![P(D)=P(A)\cdot P(D/A)+P(B)\cdot P(D/B)+P(C)\cdot P(D/C)\\P(D)=(0.45\times 0.06)+(0.30\times 0.07)+(0.25\times 0.055)\\P(D)=0.027+0.021+0.01375\\P(D)=0.06175\ or\ P(D)=0.06175\times 100\approx 6.2\%](https://tex.z-dn.net/?f=P%28D%29%3DP%28A%29%5Ccdot%20P%28D%2FA%29%2BP%28B%29%5Ccdot%20P%28D%2FB%29%2BP%28C%29%5Ccdot%20P%28D%2FC%29%5C%5CP%28D%29%3D%280.45%5Ctimes%200.06%29%2B%280.30%5Ctimes%200.07%29%2B%280.25%5Ctimes%200.055%29%5C%5CP%28D%29%3D0.027%2B0.021%2B0.01375%5C%5CP%28D%29%3D0.06175%5C%20or%5C%20P%28D%29%3D0.06175%5Ctimes%20100%5Capprox%206.2%5C%25)
Therefore, the probability that a randomly selected automobile manufactured by the company will have a defective microprocessor is 6.2 %.