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Sagot :
a. The probability of defective shells from the given samples are:
a. Probability of zero defective shells of samples is equal to 0.9510.
b. Probability of one defective shells of samples is equal to 0.04803
c. Probability of more than 1 defective shells of samples is equal to 0.00107.
As given in the question,
Total number of batches = 350
Sample size used to test for each batch 'n' = 5
Actual defective shells (success) 'p' = 1%
= 0.01
Actual non-defective shells ' 1 - p ' = 1 - 0.01
= 0.99
a. Probability of zero defective shells from the given samples is
= ⁵C₀ × ( 0.01 )⁰ × (0.99)⁵⁻⁰
= [ (5!)/(5-0)!0! ] × 1 × ( 0.99)⁵
= 0.9510
b. Probability of 1 defective shells from the given samples is
= ⁵C₁ × ( 0.01 )¹ × (0.99)⁵⁻¹
= [ (5!)/(5-1)!1! ] × 0.01 × ( 0.99)⁴
= 5 × 0.01 × 0.9606
=0.04803
c. Probability of more than 1 defective shells from the given samples is
= P(2) + P(3) + P(4) + P(5)
= ⁵C₂ × ( 0.01 )²× (0.99)⁵⁻² + ⁵C₃ × ( 0.01 )³ × (0.99)⁵⁻³ + ⁵C₄× ( 0.01 )⁴ × (0.99)⁵⁻⁴+ ⁵C₅× ( 0.01 )⁵× (0.99)⁵⁻⁵
= 0.00097 + 0.000098 +0.000000044 +0.0000000001
=0.00107
Therefore, the probability of following conditions are as follow:
a. Probability of zero defective shells of samples is equal to 0.9510.
b. Probability of one defective shells of samples is equal to 0.04803
c. Probability of more than 1 defective shells of samples is equal to 0.00107.
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