A quality control inspector has drawn a sample of 19 light bulbs from a recent production lot. If the number of defective bulbs is 1 or less, the lot passes inspection. Suppose 30% of the bulbs in the lot are defective. What is the probability that the lot will pass inspection

0.0016283

Step-by-step explanation:

Given that:

Proportion of defective bulbs, p = 30% = 0.3

Sample size, n = 19 bulbs

Probability that the lot will pass inspection :

P(none of the 19 is defective) Or P(only one of the 19 is defective)

P(none of the 19 is defective) = (1 - p) ^n = (1 - 0.3)^19 ; 0.7^19

0.7^19 = 0.0011398

P(only one of the 19 is defective) :

P(1 defective) * P(18 not defective )

(0.3) * (1 - 0.3)^18

0.3 * 0.7^18

0.3 * 0.001628413597910449 = 0.0004885

Hence,

P(none of the 19 is defective) + P(only one of the 19 is defective)

0.0011398 + 0.0004885) = 0.0016283

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step-by-step explanation: