Question

A federal report found that a lie detector test given to someone telling the truth will suggest that they lied in about 20% of cases. A company asks the 11 job applicants for a new position about thefts from previous employers. (a) What is the probability that the lie detector indicates that at least one of the applicants is lying

Answer 1

Answer:

$0.055$

Step-by-step explanation:

Let $p$ be the probability of lying an applicant, so,  $p=20 \%$.

$\Rightarrow p=\frac {20}{100}=\frac {1}{5}\;\cdots (i)$

Any applicant will either lie or will tell the truth.

Let $q$ be the probability of telling the truth, so, $q= 1-\frac {1}{5}$

$\Rightarrow q=\frac {4}{5}$

As for every applicant $p+q=1$, so this is Bernoullies trials, for which

the probability of success of exactly $x$ times of an event out of $n$ trials is $P(x)=\binom {n}{x} p^xq^{n-x}\; \cdots (iii)$.

Now, let $E$ be the event of at least one of the applicants is lying out of $11$ applicants, here the total number of applicants, $n=11$.

So, $P(E)= P(x=1)+ P(x=2)+\cdots+ P(x=11)$

This is equivalent to

$P(E)=1-P(x=0)$ as$[P(x=0)+P(x=1)+ P(x=2)+\cdots+ P(x=11)=1]$

Now, from equation (iii),

$P(E)=1-\binom {11}{0} p^0q^{11-0}$

$\Rightarrow P(E)=1-11\left(\frac {1}{5}\right)^0\left(\frac {4}{5}\right)^{11}$   [from equations s (i) and (ii)]

$\Rightarrow P(E)=1-11\times 1 \times \left(\frac {4}{5}\right)^{11}$

$\Rightarrow P(E)=1-0.945$ (approx)

$\Rightarrow P(E)=0.055$

Hence, the probability that the lie detector indicates that at least one of the applicants is lying is $0.055.$