Answer:
[tex]a. \ P(X=5)=0.2417\\\\b. \ P(X\geq 6)=0.3056\\\\c. \ P(X<3)=0.2255[/tex]
Step-by-step explanation:
a. This is a binomial probability distribution problem expressed as:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}[/tex]
Where:
- [tex]p[/tex] is the probability a successful event.
- [tex]x[/tex] the number of successful events.
- [tex]n[/tex] is the total number of events.
Given n=10, p=0.47 the probability of exactly 5 is calculated as:
[tex]P(X=5)={n\choose x}p^x(1-p)^{n-x}\\\\={10\choose 5}0.47^5(1-0.47)^5\\\\=0.2417[/tex]
Hence, the probability that exactly 5 adults have very little confidence is 0.2417
b. Given n=10, p=0.47,
-We substitute our values in formula and the probability of at least 6 is calculated as:
[tex]P(X\geq 6)={n\choose x}p^x(1-p)^{n-x}\\\\\\\ \ \ \ \ \ ={10\choose 6}0.47^6(0.53)^4+{10\choose 7}0.47^7(0.53)^3+{10\choose 8}0.47^8(0.53)^2+{10\choose 9}0.47^9(0.53)^1+{10\choose 10}0.47^{10}(0.53)^0\\\\\\=0.1786+0.0905+0.0301+0.0059+0.0005\\\\\\=0.3056[/tex]
Hence, the probability of at least 6 adults is 0.3056
c. The probability of of less that four adults is calculated as:
[tex]P(X<4)={n\choose x}p^x(1-p)^{n-x}\\\\\\\ \ \ \ \ \ ={10\choose 0}0.47^0(0.53)^{10}+{10\choose 1}0.47^1(0.53)^9+{10\choose 2}0.47^2(0.53)^8+{10\choose 3}0.47^3(0.53)^7\\\\\\=0.0017+0.0155+0.0619+0.1464\\\\\\=0.2255[/tex]
Hence, the probability that less that 4 adults have confidence in the newspapers is 0.2255
