housing units are randomly selected ,find:
A.the probability that all of the residents have an unlisted number>?
b.the probabilty that at leats one of the numbers out of the 4 is unlisted.
Probability that all of the residents have an unlisted number?
= 0.2 * 0.2 * 0.2 * 0.2 * 0.2
= (0.2)^5
= 0.00032 or 0.032%
Probabilty that at least one of the numbers out of the 4 is unlisted.
Answer: Find the probability that all of the numbers are listed
and subtract that from 1.00:
P(all numbers are listed) = (0.8)^5 = 0.32768
1 – 0.32768 = .67232 or 67.2%
Good luck in your studies,
~ Mitch ~