Question

in a lie detector​ test, 65% of lies are identified as lies and that​ 14% of true statements are also identified as lies. 93% of the job applicants tell the truth during the polygraph test. What is the probability that a person who fails the test was actually telling the​ truth?

Answer 1

Answer:

Given that the person fails the test, the probability that person was telling the truth would be  approximately $0.74$ (or equivalently, $74\,\%$.)  

Step-by-step explanation:

Start by drawing a probability tree. What the person said would affect the outcome of the test. As a result, the first split in this tree should be based on whether the person told the truth or not.

$\left\lbrace\begin{aligned}& \text{Truth} && (0.93) \\ & \text{Lie} && (1 - 0.93)\end{aligned}\right.$.

Assume that the person told the truth, there's $0.14$ probability that the person might still fail the test.

$\left\lbrace\begin{aligned}& \text{Truth} && (0.93) \left\lbrace\begin{aligned}& \text{Pass} && (1 - 0.14) \\& \text{Fail} && (0.14)\end{aligned} \right.\\ & \text{Lie} && (1 - 0.93)\end{aligned}\right.$.

Similarly, assume that the person told the lie, there's a $0.65$ probability that the person would fail the test.

$\left\lbrace\begin{aligned}& \text{Truth}\quad (0.93) && \left\lbrace \begin{aligned}& \text{Pass} && (1 - 0.14) \\& \text{Fail} && (0.14)\end{aligned} \right.\\ & \text{Lie} \quad (1 - 0.93) && \left\lbrace\begin{aligned}& \text{Pass} && (1 - 0.65) \\& \text{Fail} && (0.65)\end{aligned} \right.\end{aligned}\right.$.

Calculate the following:

• $P(\text{Truth} \cap \text{Pass})$.
• $P(\text{Truth} \cap \text{Fail})$.
• $P(\text{Lie} \cap \text{Pass})$.
• $P(\text{Lie} \cap \text{Fail})$.

For example, here's how find $P(\text{Truth} \cap \text{Pass})$ using the tree. (That's the probability that the person told the truth and passed the test.)

Start from the Truth-Lie split at the left-hand side. In $P(\text{Truth} \cap \text{Pass})$, the person told the truth. Hence, choose the "Truth" branch. The probability of that branch is $0.93$. Continue to the second branch Pass-Fail branch. In

There are two probabilities along that path: $0.93$ and $(1 - 0.14)$. $P(\text{Truth} \cap \text{Pass})$is the same as the probability of taking that path. Therefore, $P(\text{Truth} \cap \text{Pass}) = 0.93 \times (1 - 0.14) = 0.7998$.

Do the same for the other branches.

• $P(\text{Truth} \cap \text{Pass}) = 0.93 \times (1 - 0.14) = 0.7998$.
• $P(\text{Truth} \cap \text{Fail}) = 0.93 \times 0.14 = 0.1302$.
• $P(\text{Lie} \cap \text{Pass}) = (1 - 0.93) \times (1 - 0.65) = 0.0245$.
• $P(\text{Lie} \cap \text{Fail}) = (1 - 0.93) \times 0.65 = 0.0455$..

The problem is asking for the conditional probability that the person fails the test given that person told the truth. That's written as $P(\text{Truth}\;| \; \text{Fail})$ (the vertical bar stands for "given.") That's a conditional probability. The formula for the conditional probability of A-given-B is:

$\displaystyle P(A\; | \; B) = \frac{P(A \cap B)}{P(B)}$.

In this case,

$\begin{aligned} & P(\text{Truth}\;| \; \text{Fail}) = \frac{P(\text{Truth}\cap \text{Fail})}{P(\text{Fail})}\end{aligned}$.

The value of $P(\text{Truth}\cap \text{Fail})$ has already been found. It would be necessary to find $P(\text{Fail})$.

That person can fail the test either after telling a truth or after telling a lie. There's no other way to fail the test. That means

$\begin{aligned}& P(\text{Fail}) \\ &= P(\text{Truth} \cap \text{Fail}) + P(\text{Lie} \cap \text{Fail}) \\ &= 0.1302 + 0.0455 = 0.1757 \end{aligned}$.

Apply the formula for conditional probabilities:

$\begin{aligned} & P(\text{Truth}\;| \; \text{Fail}) \\ &= \frac{P(\text{Truth}\cap \text{Fail})}{P(\text{Fail})} \\ &= \frac{0.7998}{0.1757} \approx 0.74\end{aligned}$.