Answer:
To solve for the probability that at least 3 tools out of 10 have defects in a given week, given that each tool has a 10% chance of having defects, you can use the binomial probability formula. This formula calculates the probability of exactly k successes (defective tools in this case) out of n trials (the 10 tools in the sample) with a given probability of success (10% or 0.10).
The probability of finding at least 3 tools with defects is the sum of the probabilities of finding 3, 4, 5, ..., 10 defective tools. Here's the calculation:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) + ... + P(X = 10)
Using the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
Where:
- n is the number of trials (10 tools).
- k is the number of successes (defective tools, which we're calculating for).
- p is the probability of success (10% or 0.10).
Now, calculate each term and sum them up:
P(X = 3) = (10 choose 3) * (0.10)^3 * (0.90)^7
P(X = 4) = (10 choose 4) * (0.10)^4 * (0.90)^6
P(X = 5) = (10 choose 5) * (0.10)^5 * (0.90)^5
...
P(X = 10) = (10 choose 10) * (0.10)^10 * (0.90)^0
Calculate each of these probabilities, and then sum them up to find P(X ≥ 3). The result should be very close to 1 (or 100%), indicating a high probability of finding at least 3 defective tools in a random sample of 10.
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