A Manufacturer of torque converters claim only 1% of them are defective.
A. In a shipment of 12 torque converters calculate the probability exactly 2 converters are defective.
B. In a shipment of 35 converters, calculate the probability at least 1 converter is defective.
To calculate these probabilities, we can use the binomial probability formula. The formula is:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes
- n is the number of trials (converters in this case)
- k is the number of successes (defective converters in this case)
- p is the probability of success in a single trial (probability of a converter being defective)
- (n C k) is the binomial coefficient, which can be calculated as n! / (k! * (n - k)!)
Now, let's calculate the probabilities:
A. In a shipment of 12 torque converters, we want to calculate the probability of exactly 2 converters being defective. Here, n = 12, k = 2, and p = 0.01 (1% defective rate).
P(X = 2) = (12 C 2) * (0.01)^2 * (1 - 0.01)^(12 - 2)
Using the binomial coefficient, (12 C 2) = 66.
P(X = 2) = 66 * (0.01)^2 * (0.99)^10
= 0.1209
Therefore, the probability of exactly 2 converters being defective is approximately 0.1209, or 12.09%.
B. In a shipment of 35 converters, we want to calculate the probability of at least 1 converter being defective. This is the complement of having all the converters be non-defective.
P(at least 1 defective) = 1 - P(no defective)
P(no defective) = (35 C 0) * (0.01)^0 * (0.99)^(35 - 0)
= 1 * 1 * 0.99^35
≈ 0.6547
Therefore, P(at least 1 defective) = 1 - 0.6547
≈ 0.3453
Hence, the probability of at least 1 converter being defective in a shipment of 35 converters is approximately 0.3453, or 34.53%.