Question

in 1970, 85% of 30 - year - olds in a country earned more than their parents did at age 30 (adjusted for inflation). in 2014, only 45% of 30 - year - olds in the same country earned more than their parents did at age 30. complete parts a to d below.
(a) what is the probability a randomly selected 30 - year - old in 1970 earned more than their parents at age 30?
the probability is 0.8500. (round to four decimal places as needed.)
(b) what is the probability that two randomly selected 30 - year - olds in 1970 earned more than their parents at age 30?
the probability is 0.7225. (round to four decimal places as needed.)
(c) what is the probability that out of ten randomly selected 30 - year - olds in 1970, at least one did not earn more than their parents at age 30?
the probability is . (round to four decimal places as needed.)

Explanation:

Step1: Find the probability that all ten earned more

The probability that a single 30 - year - old in 1970 earned more than their parents is $p = 0.85$. The probability that all ten randomly selected 30 - year - olds earned more than their parents is $P(X = 10)=0.85^{10}$ according to the multiplication rule for independent events.
$0.85^{10}\approx0.1969$

Step2: Find the probability that at least one did not earn more

The probability that at least one did not earn more than their parents is the complement of the event that all ten earned more. Let $A$ be the event that at least one did not earn more. Then $P(A)=1 - P(X = 10)$.
$P(A)=1 - 0.1969 = 0.8031$

Answer:

$0.8031$