13 A company studied the number of lost time acedents occurring at its Brownsville, Texas, plant. Historical records show that 0% of the employees suffered lost
time accidents last yean Management believes that a special safety program will reduce such accidents to 6% during the current year. In addition, it estimates
that 10% of employees who had lost time accidents last year will experience a lost time accident during the current yean
a. What percentage of the employees will experience lost-time accidents in both years (to 2 decimale)
0.015
b. What percentage of the employees will suffer at least one lost time accident over the two year period (to 2 decimals)?
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A) 0.01

B) 0.07

Step-by-step explanation:

We are given;

P(accidents last year) = 0% = 0

P(accidents this year) = 6% = 0.06

P(accidents last year | accidents this year) = 10% = 0.1

A) We want to find the percentage of the employees will experience lost-time accidents in both years.

From multiplication rule in probability, we know that;

P(A & B) = P(A) × P(B|A) = P(B) × P(A|B)

Thus;

P(accident in both years) = P(accidents this year) × P(accidents last year | accidents this year)

P(accident in both years) = 0.06 × 0.1

P(accident in both years) = 0.006

To 2 decimal places gives; 0.01

B) from the addition rule, we know that;

P(A or B) = P(A) + P(B) - P(A & B)

Thus;

P(accident last year or this year) = P(accidents last year) + P(accidents this year) - P(accident in both years)

P(accident last year or this year) = 0 + 0.06 + 0.006 = 0.066

To 2 decimal places = 0.07

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