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A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is
- a)0.0036
- b)0.1937
- c)0.2234
- d)0.3874
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A lot has 10% defective items. Ten items are chosen randomly from this...
Understanding the Problem
In this scenario, we have a lot of items with 10% defective rate. We are tasked with finding the probability that exactly 2 out of 10 randomly chosen items are defective.
Key Concepts
- Defective Rate: 10% (0.1)
- Non-defective Rate: 90% (0.9)
- Total Items Chosen: 10
- Defective Items Required: 2
Using the Binomial Probability Formula
The probability of exactly k successes (defective items in this case) in n trials is given by:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n = total number of trials (10)
- k = number of successes (2)
- p = probability of success (0.1)
- (n choose k) = n! / [k!(n-k)!]
Calculating the Probability
1. Calculate (10 choose 2):
- (10 choose 2) = 10! / [2!(10-2)!] = 45
2. Calculate p^k:
- p^2 = (0.1)^2 = 0.01
3. Calculate (1-p)^(n-k):
- (0.9)^(10-2) = (0.9)^8 ≈ 0.43046721
4. Combine the Values:
- P(X = 2) = 45 * 0.01 * 0.43046721 ≈ 0.1937
Conclusion
Thus, the probability that exactly 2 of the chosen items are defective is approximately 0.1937, which aligns with option B.
In this scenario, we have a lot of items with 10% defective rate. We are tasked with finding the probability that exactly 2 out of 10 randomly chosen items are defective.
Key Concepts
- Defective Rate: 10% (0.1)
- Non-defective Rate: 90% (0.9)
- Total Items Chosen: 10
- Defective Items Required: 2
Using the Binomial Probability Formula
The probability of exactly k successes (defective items in this case) in n trials is given by:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n = total number of trials (10)
- k = number of successes (2)
- p = probability of success (0.1)
- (n choose k) = n! / [k!(n-k)!]
Calculating the Probability
1. Calculate (10 choose 2):
- (10 choose 2) = 10! / [2!(10-2)!] = 45
2. Calculate p^k:
- p^2 = (0.1)^2 = 0.01
3. Calculate (1-p)^(n-k):
- (0.9)^(10-2) = (0.9)^8 ≈ 0.43046721
4. Combine the Values:
- P(X = 2) = 45 * 0.01 * 0.43046721 ≈ 0.1937
Conclusion
Thus, the probability that exactly 2 of the chosen items are defective is approximately 0.1937, which aligns with option B.
Community Answer
A lot has 10% defective items. Ten items are chosen randomly from this...
Concept:
The probability that exactly 2 of the chosen items are defective is given as,
By Binomial distribution,
where, p = probability of success, q = probability of failure
Calculation:
Given:
n = 10, x = 2, p = 0.1, q = 0.9
Therefore, P(exactly 2 of the chosen items are defective) = 10C2 × 0.12 × 0.910 - 2 ⇒ 45 × 0.01 × 0.43 = 0.1937
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A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective isa)0.0036b)0.1937c)0.2234d)0.3874Correct answer is option 'B'. Can you explain this answer?
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