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The probability that a part manufactured by a company will be defective is 0.05. If 15 such parts
are selected randomly and inspected, then the probability that at least two parts will be
defective is __________ (round off to two decimal places).
are selected randomly and inspected, then the probability that at least two parts will be
defective is __________ (round off to two decimal places).
Correct answer is '0.17'. Can you explain this answer?
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Verified Answer
The probability that a part manufactured by a company will be defectiv...
Total no. of parts = 15
P[defective(p)] = 0.05
P[non-defective(q)] = 1 – 0.05 = 0.95
P[at least 2 defective] = 1 – [P(no defective)+P(1 defective)]
Using the binomial distribution,
P[at least 2 defective] = 1 – [15C0p0q15 + 15C1p1q14]= 1 – [0.0500.9515 + 15× 0.05 × 0.9514] = 1 – [0.4639 + 0.3657] = 1 – 0.8296 = 0.1704
P[defective(p)] = 0.05
P[non-defective(q)] = 1 – 0.05 = 0.95
P[at least 2 defective] = 1 – [P(no defective)+P(1 defective)]
Using the binomial distribution,
P[at least 2 defective] = 1 – [15C0p0q15 + 15C1p1q14]= 1 – [0.0500.9515 + 15× 0.05 × 0.9514] = 1 – [0.4639 + 0.3657] = 1 – 0.8296 = 0.1704
Most Upvoted Answer
The probability that a part manufactured by a company will be defectiv...
Understanding the Problem
To find the probability of at least two defective parts out of 15, we start with the given defect probability and use the complementary probability approach.
Given Information
- Probability of a defective part (p) = 0.05
- Probability of a non-defective part (q) = 1 - p = 0.95
- Total parts selected (n) = 15
Complementary Probability Approach
1. Calculate the probabilities for 0 and 1 defective parts:
- The probability of 0 defective parts (X = 0) can be calculated using the binomial formula:
P(X = 0) = (n choose 0) * (p^0) * (q^n) = (1) * (0.05^0) * (0.95^15)
- The probability of 1 defective part (X = 1):
P(X = 1) = (n choose 1) * (p^1) * (q^(n-1)) = (15 choose 1) * (0.05^1) * (0.95^14)
2. Calculate the probabilities:
- P(X = 0) = 0.95^15 ≈ 0.4633
- P(X = 1) = 15 * (0.05) * (0.95^14) ≈ 0.3670
3. Combine these probabilities:
- Total probability of 0 or 1 defective part:
P(X ≤ 1) = P(X = 0) + P(X = 1) ≈ 0.4633 + 0.3670 ≈ 0.8303
Final Calculation
To find the probability of at least two defective parts:
- P(X ≥ 2) = 1 - P(X ≤ 1) ≈ 1 - 0.8303 ≈ 0.1697
Rounding this off to two decimal places gives us 0.17.
Conclusion
Thus, the probability that at least two parts will be defective is approximately 0.17.
To find the probability of at least two defective parts out of 15, we start with the given defect probability and use the complementary probability approach.
Given Information
- Probability of a defective part (p) = 0.05
- Probability of a non-defective part (q) = 1 - p = 0.95
- Total parts selected (n) = 15
Complementary Probability Approach
1. Calculate the probabilities for 0 and 1 defective parts:
- The probability of 0 defective parts (X = 0) can be calculated using the binomial formula:
P(X = 0) = (n choose 0) * (p^0) * (q^n) = (1) * (0.05^0) * (0.95^15)
- The probability of 1 defective part (X = 1):
P(X = 1) = (n choose 1) * (p^1) * (q^(n-1)) = (15 choose 1) * (0.05^1) * (0.95^14)
2. Calculate the probabilities:
- P(X = 0) = 0.95^15 ≈ 0.4633
- P(X = 1) = 15 * (0.05) * (0.95^14) ≈ 0.3670
3. Combine these probabilities:
- Total probability of 0 or 1 defective part:
P(X ≤ 1) = P(X = 0) + P(X = 1) ≈ 0.4633 + 0.3670 ≈ 0.8303
Final Calculation
To find the probability of at least two defective parts:
- P(X ≥ 2) = 1 - P(X ≤ 1) ≈ 1 - 0.8303 ≈ 0.1697
Rounding this off to two decimal places gives us 0.17.
Conclusion
Thus, the probability that at least two parts will be defective is approximately 0.17.
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The probability that a part manufactured by a company will be defective is 0.05. If 15 such partsare selected randomly and inspected, then the probability that at least two parts will bedefective is __________ (round off to two decimal places).Correct answer is '0.17'. Can you explain this answer?
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