Respuesta :
Answer:
a) [tex]P(X=5)=0.208[/tex]
b) [tex]P(X\geq 6)=0.1854[/tex]
c) [tex]P(X<4)=0.3566[/tex]
Step-by-step explanation:
Given : 41% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan.
To Find : The probability that the number who consider themselves baseball fans is (a) exactly five, (b) at least six, and (c) less than four.
Solution :
Applying binomial theorem,
[tex]P(X=r)=^nC_r\times p^r\times (1-p)^{n-r}[/tex]
The success p= 41%=0.41
The failure (1-p)=1-041=0.59
Number of selection n=10
a) The probability that the number who consider themselves baseball fans is exactly five,
i.e. X=5
[tex]P(X=5)=^{10}C_5\times (0.41)^5\times (0.59)^{10-5}[/tex]
[tex]P(X=5)=\frac{10!}{5!5!}\times (0.41)^5\times (0.59)^{5}[/tex]
[tex]P(X=5)=252\times 0.0115\times 0.071[/tex]
[tex]P(X=5)=0.208[/tex]
b) The probability that the number who consider themselves baseball fans is at least six,
i.e. [tex]X\geq 6[/tex]
[tex]P(X\geq 6)=1-P(X<6)[/tex]
[tex]P(X\geq 6)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)][/tex]
[tex]P(X\geq 6)=1-[^{10}C_0\times (0.41)^0\times (0.59)^{10-0}+^{10}C_1\times (0.41)^1\times (0.59)^{10-1}+^{10}C_2\times (0.41)^2\times (0.59)^{10-2}+^{10}C_3\times (0.41)^3\times (0.59)^{10-3}+^{10}C_4\times (0.41)^4\times (0.59)^{10-4}+^{10}C_5\times (0.41)^5\times (0.59)^{10-5}][/tex]
[tex]P(X\geq 6)=1-[0.0051+0.0355+0.111+0.205+0.250+0.208][/tex]
[tex]P(X\geq 6)=1-[0.8146][/tex]
[tex]P(X\geq 6)=0.1854[/tex]
c) The probability that the number who consider themselves baseball fans is less than four,
i.e. [tex]X<4[/tex]
[tex]P(X<4)=P(=0)+P(X=1)+P(X=2)+P(X=3)[/tex]
[tex]P(X<4)=^{10}C_0\times (0.41)^0\times (0.59)^{10-0}+^{10}C_1\times (0.41)^1\times (0.59)^{10-1}+^{10}C_2\times (0.41)^2\times (0.59)^{10-2}+^{10}C_3\times (0.41)^3\times (0.59)^{10-3}[/tex]
[tex]P(X<4)=0.0051+0.0355+0.111+0.205[/tex]
[tex]P(X<4)=0.3566[/tex]
