On-the-job arrogance and task performance. Human Performance (Vol. 23, 2010) published the results of a study that found that arrogant workers are more likely to have poor performance ratings. Suppose that 15% of all full-time workers exhibit arrogant behaviors on the job and that 10% of all full-time workers will receive a poor performance rating. Also, assume that 5% of all full-time workers exhibit arrogant behaviors and receive a poor performance rating. Let A be the event that a full-time worker exhibits arrogant behavior. Let B be the event that a full-time worker will receive a poor performance rating.

a. Are the events A and B mutually exclusive? Explain.

b. Find P(B/A).

c. Are the events A and B independent? Explain.

In a probability experiment, if an event A and B are two independent occurrences, the chance that both events occur at the same time is:

$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = P (A) x P (B)}}$

In the case of dependent events, the likelihood that both occurrences occur concurrently is:

$\mathbf{\text{P (B/A) =}}\frac{\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B)}}}{\mathbf{\text{P (A)}}}$

Here, we have

$\begin{array}{c}\mathbf{\text{P (A) = 15\%}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{15}}}{\mathbf{\text{100}}}\\ \mathbf{\text{=0.15}}\end{array}$

$\begin{array}{c}\mathbf{\text{P (B) = 10\%}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{10}}}{\mathbf{\text{100}}}\\ \mathbf{\text{=0.1}}\end{array}$

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B)= 5\%}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{5}}}{\mathbf{\text{100}}}\\ \mathbf{\text{=0.05}}\end{array}$

Only one of the two occurrences may occur at the same time for the two events to be mutually exclusive, which indicates that:

$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0}}$

Therefore, the two events are not mutually exclusive.

$\begin{array}{c}\mathbf{\text{P (B/A) =}}\frac{\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B)}}}{\mathbf{\text{P (A)}}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{0.05}}}{\mathbf{\text{0.15}}}\\ \mathbf{\text{= 0.34}}\end{array}$

Hence, the required probability is0.34.

To be independent, the chance of one event occurring does not impact the probability of the other, which indicates that:

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = P (A)}}\mathbf{\times }\mathbf{\text{P (B)}}\\ \mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0.15}}\mathbf{\times }\mathbf{\text{0.1}}\\ \mathbf{\text{= 0.015}}\end{array}$

Therefore, the two events are not independent.