Question

59% 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)=0.209 (round to three decimal places as needed.) (b) p(x≥6)=0.608 (round to three decimal places as needed.) (c) p(x<4)=□ (round to three decimal places as needed.)

Explanation:

Step1: Assume this is a binomial - probability problem

Let \(n = 10\) (number of trials, i.e., number of U.S. adults selected), \(p=0.59\) (probability that a U.S. adult has very little confidence in newspapers). The binomial - probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Calculate \(P(X\lt4)\)

\(P(X\lt4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)
For \(k = 0\):
\(C(10,0)=\frac{10!}{0!(10 - 0)!}=1\), \(P(X = 0)=C(10,0)\times(0.59)^{0}\times(1 - 0.59)^{10}=1\times1\times(0.41)^{10}\approx0.0001\)
For \(k = 1\):
\(C(10,1)=\frac{10!}{1!(10 - 1)!}=10\), \(P(X = 1)=C(10,1)\times(0.59)^{1}\times(0.41)^{9}=10\times0.59\times(0.41)^{9}\approx0.0016\)
For \(k = 2\):
\(C(10,2)=\frac{10!}{2!(10 - 2)!}=45\), \(P(X = 2)=C(10,2)\times(0.59)^{2}\times(0.41)^{8}=45\times0.59^{2}\times(0.41)^{8}\approx0.0093\)
For \(k = 3\):
\(C(10,3)=\frac{10!}{3!(10 - 3)!}=120\), \(P(X = 3)=C(10,3)\times(0.59)^{3}\times(0.41)^{7}=120\times0.59^{3}\times(0.41)^{7}\approx0.0337\)
\(P(X\lt4)=0.0001 + 0.0016+0.0093 + 0.0337=0.0447\approx0.045\)

Answer:

(a) \(P(5)=0.209\)
(b) \(P(X\geq6)=0.608\)
(c) \(P(X\lt4)=0.045\)