Respuesta :
Answer:
[tex]0.055[/tex]
Step-by-step explanation:
Let [tex]p[/tex] be the probability of lying an applicant, so, [tex]p=20 \%[/tex].
[tex]\Rightarrow p=\frac {20}{100}=\frac {1}{5}\;\cdots (i)[/tex]
Any applicant will either lie or will tell the truth.
Let [tex]q[/tex] be the probability of telling the truth, so, [tex]q= 1-\frac {1}{5}[/tex]
[tex]\Rightarrow q=\frac {4}{5}[/tex]
As for every applicant [tex]p+q=1[/tex], so this is Bernoullies trials, for which
the probability of success of exactly [tex]x[/tex] times of an event out of [tex]n[/tex] trials is [tex]P(x)=\binom {n}{x} p^xq^{n-x}\; \cdots (iii)[/tex].
Now, let [tex]E[/tex] be the event of at least one of the applicants is lying out of [tex]11[/tex] applicants, here the total number of applicants, [tex]n=11[/tex].
So, [tex]P(E)= P(x=1)+ P(x=2)+\cdots+ P(x=11)[/tex]
This is equivalent to
[tex]P(E)=1-P(x=0)[/tex] as[tex][P(x=0)+P(x=1)+ P(x=2)+\cdots+ P(x=11)=1][/tex]
Now, from equation (iii),
[tex]P(E)=1-\binom {11}{0} p^0q^{11-0}[/tex]
[tex]\Rightarrow P(E)=1-11\left(\frac {1}{5}\right)^0\left(\frac {4}{5}\right)^{11}[/tex] [from equations s (i) and (ii)]
[tex]\Rightarrow P(E)=1-11\times 1 \times \left(\frac {4}{5}\right)^{11}[/tex]
[tex]\Rightarrow P(E)=1-0.945[/tex] (approx)
[tex]\Rightarrow P(E)=0.055[/tex]
Hence, the probability that the lie detector indicates that at least one of the applicants is lying is [tex]0.055.[/tex]
