Respuesta :
Let the probability of a brake defect be P(B), the probability of a chain defect be P(C) then the probability of a brake and a chain defect is P(B ∩ C) and the probability of a brake or a chain defect P(B ∪ C).
Given that P(B) = 4%, P(C) = ?, P(B ∩ C) = 1%, P(B ∪ C) = 6%.
P(B ∪ C) = P(B) + P(C) - P(B ∩ C)
Thus, P(C) = P(B ∪ C) + P(B ∩ C) - P(B) = 6% + 1% - 4% = 3%
Given that P(B) = 4%, P(C) = ?, P(B ∩ C) = 1%, P(B ∪ C) = 6%.
P(B ∪ C) = P(B) + P(C) - P(B ∩ C)
Thus, P(C) = P(B ∪ C) + P(B ∩ C) - P(B) = 6% + 1% - 4% = 3%
If the probability of a defect with the brakes or the chain is 6 percent, the probability of a chain defect is 3%.
Probability
Using this formula
P(C) = P(B ∪ C) + P(B ∩ C) - P(B)
Where:
Brake defect= P(B) = 4%
Chain defect= P(C) = ?
Brake defect and chain defect =P(B ∩ C) = 1%
Brake defect or chain defect=P(B ∪ C) = 6%
Let plug in the formula
P(C) = 6% + 1% - 4%
P(C)= 3%
Therefore the probability of a chain defect is 3%.
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