Need help finding the probability for these three questions - Due in less than 4 hours. Please help

[mcsweetlez]

53% 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 confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four:

(a) P(5) = ? (Round to three decimal places as needed.)
(b) P(x6) = ? (Round to three decimal places as needed.)
(c) P(x4) = ? (Round to three decimal places as needed.)

 

[R.M.]

A least 90% of the people that regularly watch these forums, and are willing to give freely of their time to help people improve their skills, would love to hear your thoughts on the problem, so that they may better help you fill in the gaps in your understanding. Based on how you worded your "question", I would wager that nearly 100% of the people that read it most likely interpreted it as an attempt to have someone do your homework for you.

 

[HallsofIvy]

It's due in four hours and you haven't yet tried anything? Don't you think you should get started?

(It's odd that this was due at 11 P,M. but since that deadline is past-
This is a binomial distribution where the "probability of success", p, is 0.53, the "probability of failure", q, is 0.47, and n= 10. The probability of i "successes" and n-i "failures" is \(\displaystyle \begin{pmatrix}10 \\ i\end{pmatrix}(0.53)^i(0.47)^{10-i}\).

For example, \(\displaystyle P(5)= \begin{pmatrix}10 \\ 5\end{pmatrix} (0.53)^5(0.47)^5=\frac{10!}{5!5!}(0.530)^5(0.47)^5= \frac{10(9)(8)(7)(6)}{5(4)(3)(2)}(0.0418195493)(0.0229345007)= 2(9)(2)(7)(0.00095911048269453451)= 0.24169584163902269652\). Rounded to three decimal places that is 0.242.)