Question

(a) Four boxes A, B, C, D contain fuses The boxes contain 5000, 3000, 2000 & 1000 fisses respectively. The percentages of fuses in the boxes which are defective are 3%, 2%, 1% and 0.3% respectively. One fuse is selected at random arbitrarily from one of the boxes. It is found to be a defective fuse. Find the probability that it has come from box D.​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Given :  Four boxes A, B, C, D contain fuses

The boxes contain 5000, 3000, 2000 & 1000 fuses respectively.

The percentages of fuses in the boxes which are defective are 3%, 2%, 1% and 0.3% respectively.

One fuse is selected at random arbitrarily from one of the boxes. It is found to be a defective fuse.

To Find : the probability that it has come from box D.​

Solution:

Defectives in Box A  = (3/100) 5000  = 150

Defectives in Box B  = (2/100) 3000  = 60

Defectives in Box C  = (1/100) 2000  = 20

Defectives in Box D  = (0.1/100) 1000  = 1

Total Defectives  = 150 + 60 + 20 + 1  = 231

probability that it has come from box D.​   = 1/231

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