A company has rated 77% of its employees as acceptable workers and 23% as unacce

Question

A company has rated 77% of its employees as acceptable workers and 23% as unacceptable. Personnel records indicate that 68% of the acceptable workers had previous work experience, while only 28% of the unacceptable workers had previous work experience. Based on the information above, if a person with pervious work experience is hired, what is the probability that this person will be a satisfactory employee? (Briefly define your events and include appropriate probability notation with your answer.) What is the probability that a randomly selected employee is an unacceptable worker but has previous work experience? (Include appropriate probability notation with your answer.)

Explanation / Answer

(a)

Percentage of employees who have previous work experience are:

0.77*0.68+0.23*0.28=0.588

This is calculated as:68% of acceptable employees have work experience so: 0.77*0.68

And 28% of unacceptable employees have work experience so:0.23*0.28

Person will be a satisfactory employee if he/she is an accetable employee

Hence probability that employee will be satisfactory is:

(0.77*0.68)/(0.77*0.68+0.23*0.28) ~ 0.89

(b)

We need that randomly selected employee is an unacceptable worker and has previous work experience

We saw before:

Percentage of employees who have previous work experience are:

0.77*0.68+0.23*0.28=0.588

Fraction contributed by unacceptable workers is: 0.23*0.28

Hence, required probability is: (0.23*0.28)/.588 ~ 0.11

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