In a large population, 61% of the people are vaccinated, meaning there are 39% who are not. The problem asks for the probability that out of the 4 randomly selected people, at least one of them has been vaccinated. Therefore, we need to add all the possibilities that there could be one, two, three or four randomly selected persons who were vaccinated.
For only one person, we use P(1), same reasoning should hold for other subscripts.
P(1) = (61/100)(39/100)(39/100)(39/100) = 0.03618459
P(2) = (61/100)(61/100)(39/100)(39/100) = 0.05659641
P(3) = (61/100)(61/100)(61/100)(39/100) = 0.08852259
P(4) = (61/100)(61/100)(61/100)(61/100) = 0.13845841
Adding these probabilities, we have 0.319761. Therefore the probability of at least one person has been vaccinated out of 4 persons randomly selected is 0.32 or 32%, rounded off to the nearest hundredths.
Answer:
(a) 0.0833 or 8.33%
(b) 0.40 or 40%
Step-by-step explanation:
Parts line one (n1) = 1,000 parts
Defects line one (d1) = 100 parts
Parts line two (n2) = 2,000 parts
Defects line two (d2) = 150 parts
Total number of parts (n) = 3,000 parts
a. Probability of a randomly selected part being defective:

The probability is 0.0833 or 8.33%
b. Probability of a part being produced by line one, given that it is defective:

The probability is 0.40 or 40%.
Answer:
Step-by-step explanation:
73-(10+3)=60 you have to add 7 and 3 first, then you subtract it from 73