SOLUTION: ​69% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confid

(a)  It is a binomial distribution probability problem.

     - number of trials        n =  10;
     - number of success trial k =   5;
     - Probability of success on a single trial p = 0.69.



We need calculate  P(n=10; k = 5; p=0.69).      


To facilitate calculations, I use an online (free of charge) calculator at this web-site 

https://stattrek.com/online-calculator/binomial.aspx


It provides nice instructions  and  a convenient input and output for all relevant options/cases.


    P(n=10; k = 5; p=0.69) = 0.11283776194,   or   0.1128 (rounded).       ANSWER




(b)  It is a binomial distribution probability problem.

     - number of trials        n =  10;
     - number of success trial k >=  6;
     - Probability of success on a single trial p = 0.69.



We need calculate  P(n=10; k >=  6; p=0.69).      


Use online (free of charge) calculator at this web-site 

https://stattrek.com/online-calculator/binomial.aspx


    P(n=10; k >= 6; p=0.69) = 0.83205254038,   or   0.8321 (rounded).       ANSWER



(c)  It is a binomial distribution probability problem.

     - number of trials        n =  10;
     - number of success trial k <   4;
     - Probability of success on a single trial p = 0.69.



We need calculate  P(n=10; k < 4; p=0.69).      


Use online (free of charge) calculator at this web-site 

https://stattrek.com/online-calculator/binomial.aspx


    P(n=10; k < 4; p=0.69) = 0.0128636757,   or   0.0129 (rounded).       ANSWER