Question

Three machines, E1, E2, E3 in a factory produce 50%, 25% and 25% respectively in a day. It is known that 4% of the total output of E1, 4% of E2 and 5% of E3 is defective. If one unit is picked up at random, then what is the probablity that it was defective. Pls answer ASAP!

Answer 1

P = (E1 and Defective) or (E2 and Defective) or (E3 and Defective)
    = 50/100 x 4/100 + 25/100 x 4/100 + 25/100 x 5/100
    =  0.0425

Answer 2

Given : Three machines, E1, E2, E3 in a factory produce 50%, 25% and 25% respectively in a day

To find : probability of  defective item

Solution:

Let say Total Production by Machine = 1000

Production by Machine E1 = (50/100) 1000 = 500

Production by Machine E2 = (25/100) 1000 = 250

Production by Machine E3 = (25/100) 1000 = 250

Defective output by E1 = (4/100) 500 = 20

Defective output by E2 = (4/100) 250 = 10

Defective output by E3 = (5/100) 250 = 12.5

Total Defectives = 20 + 10 + 12.5  = 42.5

Probability that items was defective = 42.5/1000

= 0.0425

= 4.25 %

0.0425 (4.25 % ) is  the probability that item was defective.

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