Question

You are studying 112 returning combat veterans with deployment related injuries. You are testing a cognitive impairment screen to detect traumatic brain injuries (TBI). There are six veterans with confirmed TBI and five of them screen positive. There are 93 veterans who do not have TBI and screen negative. There are a total 18 veterans who screen positive. One of the veterans has a negative screen and wants to know the probability that he does not have a TBI. You tell him:_________

Answer 1

Answer:

0.9894 = 98.94% probability that he does not have a TBI.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Negative screen

Event B: Does not have a TBI.

Probability of a negative screen:

93 are negative and do not have a TBI.

1 is negative and has a TBI.

Out of 112.

So

$P(A) = \frac{93+1}{112} = \frac{94}{112}$

Probability of a negative screen and not having a TBI:

93 are negative and do not have a TBI, out of 112, so:

$P(A \cap B) = \frac{93}{112}$

One of the veterans has a negative screen and wants to know the probability that he does not have a TBI.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{93}{112}}{\frac{94}{112}} = \frac{93}{94} = 0.9894$

0.9894 = 98.94% probability that he does not have a TBI.