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 46% of 30 - year - olds in the same country earned more than their parents did at age 30. complete parts a to d below.
(a) 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 0.8031. (round to four decimal places as needed.)
(d) what is the probability that out of ten randomly selected 30 - year - olds in 2014, 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 probability of an individual not earning more

In 2014, the probability that a 30 - year - old earned more than their parents is $p = 0.46$. So the probability that a 30 - year - old did not earn more than their parents is $q=1 - p=1 - 0.46 = 0.54$.

Step2: Find probability of all earning more

The probability that all ten randomly selected 30 - year - olds earned more than their parents is $P(X = 10)=p^{10}=(0.46)^{10}$.

Step3: Find probability of at least one not earning more

The probability that at least one did not earn more than their parents is the complement of all of them earning more. Let $A$ be the event that at least one did not earn more. Then $P(A)=1 - P(X = 10)=1-(0.46)^{10}$.
$1-(0.46)^{10}=1 - 0.000454=0.999546\approx0.9995$

Answer:

$0.9995$