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

`0.055`

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* 1 * \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.`