A factory produces components of which 10% are defective. Components are packed in boxes of 20. A box is selected at random. What is the probability that the box contains exactly 2 defective components?Question 10Select one:a.0.285b.0.811c.0.715d.0.189
Question
A factory produces components of which 10% are defective. Components are packed in boxes of 20. A box is selected at random. What is the probability that the box contains exactly 2 defective components?Question 10Select one:a.0.285b.0.811c.0.715d.0.189
Solution
To solve this problem, we can use the binomial probability formula, which 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
- n is the number of trials
- k is the number of successes
In this case:
- n = 20 (the number of components in a box)
- k = 2 (the number of defective components we're interested in)
- p = 0.10 (the probability of a component being defective)
So, we can plug these values into the formula:
P(X=2) = C(20, 2) * (0.10^2) * ((1-0.10)^(20-2))
First, calculate C(20, 2), which is the number of ways to choose 2 items from 20:
C(20, 2) = 20! / [(20-2)! * 2!] = 190
Next, calculate (0.10^2) * ((1-0.10)^(20-2)):
(0.10^2) * ((1-0.10)^(20-2)) = 0.01 * (0.90^18) = 0.0387
Finally, multiply these two results together:
P(X=2) = 190 * 0.0387 = 0.7353
So, the probability that a randomly selected box contains exactly 2 defective components is approximately 0.7353, or 73.53%. This is not an option in the given choices, so there may be a mistake in the problem or the provided options.
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