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In a police interrogation it is known that 40% of the subjects lie and the rest tell the truth; when a subject lies the probability that a polygraph will spot that lie is 0.9; and when the subject is truthful, the probability of the machine to confirm that is 0.85. With this information given, determine:

a. A subject is interrogated by the machine, what is the probability that he/she really is lying?

b. The polygraph has determined that the subject lies, what is the probability that this is really true?

c. 4 unrelated subjects are interrogated by the machine; what is the probability that the machine will determine that exactly 3 of them are lying?

The attempt at the solution:

a. 0.4*0.9 + 0.4*0.1

b. 0.4*0.9 / (0.4*0.9 + 0.6*0.1 + 0.6 *0.15)

c. From the Bernoulli theorem: ${4 \choose 3} * (0.4*0.9 + 0.6*0.1 + 0.6 *0.15)^3 *(0.49)^1 $

Is this right?

Official answers state: a. 0.45, b. 0.8 , c. 0.2 .

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(a) yes, but a quicker way is to use "it is known that 40% of the subjects lie" to give $0.4$ immediately

(b) not quite: the $0.6 \times 0.1$ should not be in the denominator

(c) same problem as (b). If the probability the polygraph determines a lie is $p=0.4\times 0.9 + 0.6 \times 0.15$ then the solution is what you used: $\displaystyle{4 \choose 3}p^3(1-p)^1$

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  • $\begingroup$ $0.6 \times 0.1$ definitely HAS to be in the denominator, since $0.6 \times 0.1$ is indeed one such case when the machine determines a lie has been given. (even if that is not true). $\endgroup$ – Bak1139 Dec 5 '13 at 18:49
  • $\begingroup$ No. In fact not a lie, but the machine determines it is a lie, has probability $0.6 \times 0.15$, which you also have in the denominator. In fact a lie, and the machine determines it is a lie, has probability $0.4 \times 0.9$, which is there too. There are no other cases where the machine determines it is a lie $\endgroup$ – Henry Dec 5 '13 at 18:54
  • $\begingroup$ $.6*.15$ has to be in the denominator as that is the chance the person tells the truth but appears to lie. I don't know what $.6*.1$ would be but it does not belong. $\endgroup$ – kaine Dec 5 '13 at 18:55
  • $\begingroup$ I see it now, yes... $\endgroup$ – Bak1139 Dec 5 '13 at 19:14

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