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To test a new car, an automobile manufacturer wants to select 4 employees to test-drive the car for 1 year. If 12 management and 8 union employees volunteer to be test drivers and the selection is made at random, what is the probability that at least 1 union employee is selected?

Answer :

The probability that at least 1 union employee is selected to test-drive the car is approximately 0.9038 or 90.38%.

To find the probability of at least 1 union employee being selected, we'll first calculate the probability of selecting only management employees for all 4 positions. We'll use combinations to find the number of ways to choose 4 management employees out of 12, divided by the total number of ways to choose 4 employees out of 20.

1. Calculate the number of ways to choose 4 management employees out of 12:

[tex]\(_{12}C_4 = \frac{12!}{4!(12-4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495\)[/tex].

2. Calculate the number of ways to choose 4 employees out of 20:

[tex]\(_{20}C_4 = \frac{20!}{4!(20-4)!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845\)[/tex].

3. Find the probability of selecting only management employees:

[tex]Probability = \(\frac{_{12}C_4}{_{20}C_4} = \frac{495}{4845} ≈ 0.1021\)[/tex].

4. Subtract this probability from 1 to find the probability of at least 1 union employee being selected:

Probability = (1 - 0.1021 ≈ 0.9038).

Therefore, the probability that at least 1 union employee is selected to test-drive the car is approximately 0.9038 or 90.38%.

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