Question

In one factory, the light bulb is produced by three machines. The first machine produces 20% of the total production of the factory. The second machine produces 30% of the total production of the factory. The third machine produces 50% of the total production of the factory. The percentage of damaged production of the first machine is 1% and the percentage of damaged production of the second machine is 4% and the percentage of damaged production of the third machine is 7%

If the lamp is chosen randomly from the production of this factory

? What is the probability that the lamp is damaged  

		0.046
		0.002
		0.012
		0.035

1 درجات   

Answer 1

9 ut A. B &c denote the event that the lamp is producted by ist, 2nd & 3rd machine respectively. If D: Lamp is defective. Given PIA)-20%= 0.2, PCB): 30%=0.3, P(c)=50% =0,5 Also P(DIA): 1%-0.01, P(DIB):0.04, PCDIC):0.07 P(Damaged): PCD)=P(PNA) +P(DNB)+P(DNC) = P(DIA) PCA) + PDB) PCB) +PCDIC) Ple) -0.01)(0.2)+(0.04)(0,3)+(0.07)(0.5) -0.049