CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In a factory, there are two machines A and B producing bulbs. Their contributions are respectively 60% and 40% of the total production. It is found that 1% and 3% of the items produced by these machines are defective. An item randomly chosen from a day's production was found to be defective. Find the probability that the defective item was produced by machine B.

Open in App
Solution

Let E1 and E2 be the events that the items produced by machine A and B respectively.
Then P(E1)=60100=35 and P(E2)=40100=25
Let C be the event that we choose a defective item at random.
Then, P(The defective item came from machine A)=P(C|E1)=1%=1100
P(The defective item came from machine B)=P(C|E2)=3%=3100
To find the probability that the defective item was produced by machine B, we need to find P(E2|C)
So, we use Baye's formula and then we have
P(E2|C)=P(E2)P(C|E2)P(E1)P(C|E1)+P(E2)P(C|E2)
=25×310035×1100+25×3100
=63+6=69=23
Hence, the probability that the defective item was produced by machine B is 23.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Conditional Probability
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon