On average, 5 % of items supplied by manufacturer X are defective. If a batch of 10 items is inspected: what is the probability that 2 items are defective 0.065 0.075 0.085 0.095
Question
On average, 5 % of items supplied by manufacturer X are defective. If a batch of 10 items is inspected: what is the probability that 2 items are defective 0.065 0.075 0.085 0.095
Solution
This is a binomial distribution problem. The formula for binomial distribution is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case:
- n = 10 (the number of items inspected)
- k = 2 (the number of defective items we're interested in)
- p = 0.05 (the probability that an item is defective)
So, we can plug these values into the formula:
P(X=2) = C(10, 2) * (0.05^2) * ((1-0.05)^(10-2))
First, calculate C(10, 2). This is the number of ways you can choose 2 items from 10, which is 45.
Next, calculate (0.05^2), which is 0.0025.
Then, calculate ((1-0.05)^(10-2)), which is (0.95^8), or approximately 0.6634.
Finally, multiply these three values together:
P(X=2) = 45 * 0.0025 * 0.6634 = 0.0746
So, the probability that 2 items are defective is approximately 0.075.
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