The probability any one widget is free of defects is 0.93. For quality control purposes the company is concerned about bad batches of widgets. If the widgets are shipped in batches of 20, find the probability of at most 2 defective widgets in one batch.
Prob 0 defective = (.93)^20
prob 1 defective = C(20,1) (.93)^19 (.07)
prob 2 defective - C(20,2) (.93)^18 (.07)^2
prob (at most 2 defective) = 1 - (sum of the above 3 probs)
To find the probability of at most 2 defective widgets in one batch, we will use the binomial distribution. The binomial distribution can be used to calculate the probability of obtaining a specific number of successes (defective widgets in this case) in a fixed number of independent Bernoulli trials (individual widgets in this case), each with the same probability of success.
The probability of success (a widget being defect-free) is given as 0.93, and the probability of failure (a widget being defective) is 1 - 0.93 = 0.07.
Let's calculate the probability using the binomial distribution formula:
P(X ≤ k) = ∑ [from i = 0 to k] (nCi * p^i * q^(n-i))
Where:
- P(X ≤ k) is the probability of obtaining at most k defective widgets in a batch.
- n is the total number of widgets in the batch (20 in this case).
- i is the number of defective widgets.
- p is the probability of success (0.93, the probability of a defect-free widget).
- q is the probability of failure (0.07, the probability of a defective widget).
- nCi is the number of ways to choose i defective widgets out of n (n choose i).
Now, let's proceed with the calculation:
P(X ≤ 2) = ∑ [from i = 0 to 2] (20Ci * 0.93^i * 0.07^(20-i))
We can use a calculator or statistical software to compute this probability. In this case, the value comes out to approximately 0.8994.
Therefore, the probability of at most 2 defective widgets in one batch is approximately 0.8994, or 89.94%.