Answer:
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
Answer: 32 is the answer
Step-by-step explanation:
Answer:it would be way over 100 because If you do the math it comes out as 312.73….. therefore it’s higher than 100 hope this helped hunny boo
