We have two boxes of fuses. Box 1 contains 10 defective and 30 non-defective fuses. Box 2 contains 5 defective and 7 non-defective fuses. A fuse is randomly chosen from box 1 and transferred to box 2. Find the probability that now a fuse chosen randomly from box 2 is defective.
Answers
Given :
Box 1 contains 10 defective and 30 non-defective fuses
Box 2 contains 5 defective and 7 non-defective fuses
A fuse is randomly chosen from box 1 and transferred to box 2
To find the probability that now a fuse chosen randomly from box2 is defective
Solution:
Box 1 contains 10 defective and 30 non-defective fuses
=> Total = 10 + 30 = 40
A fuse is randomly chosen from box 1 and transferred to box 2
Case 1 : Defective fuse moved to Box 2
probability of moving Defective fuse = 10/40 = 1/4
Then Box 2 contains 5 + 1 = 6 defective and 7 non-defective fuses
Total = 6 + 7 = 13
Selecting Defective fuse from Box 2 = 6/13
Hence probability in case 1
= (1/4)(6/13)
= 3/26
Case 2 : non-Defective fuses moved to Box 2
probability of moving non Defective fuses = 30/40 = 3/4
Then Box 2 contains 5 defective and 7+1 = 8 non-defective fuses
Total = 5 + 8 = 13
Selecting Defective fuse from Box 2 = 5/13
Hence probability in case 2
= (3/4)(5/13)
= 15/52
probability that now a fuse chosen randomly from box 2 is defective
= 3/26 + 15/52
= 21/52
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