A bag contains a certain number of bolts out of which some are 'defective' while the remaining are 'non-defective'. The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag. If 40 more 'non- defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%. Now two bolts are picked from the bag then what is the probability that at most one of the two bolts is 'defective'?

Question: A bag contains a certain number of bolts out of which some are 'defective' while the remaining are 'non-defective'. The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag. If 40 more 'non- defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%. Now two bolts are picked from the bag then what is the probability that at most one of the two bolts is 'defective'?

Solution:
Let the number of non-defective bolts in the bag = N
The number of defective bolts in the bag = D
∴ The total number of bolts in the bag = N + D

The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag.
ATQ,
N/(N + D) = 300% × D/(N + D)
∴ N = 3D ..............(i)

If 40 more 'non-defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%.
D/(N + D + 40) = 20/100
⇒ D/(3D+ D+ 40) = 1/5 [Using (i)]
⇒ 5D = 4D + 40
∴ D = 40
∴ N = 3D = 120

The total number of bolts in the bag = 40 + 120 = 160

Two bolts are picked from the bag then the probability that at most one of the two bolts is 'defective' = 1 - P (Both the selected bolts are defective)

∴ Required probability = 1 - (⁴⁰C₂/ ¹⁶⁰C₂)
= 1 - (780/12720)
= 1 - (78/1272)
= 1 - (13/212)
= (212 - 13)/212
= 199/212
Hence, the correct answer is 199/212.