Question

There are 50 workers onsite at a large construction project. Six of them are left-handed, and the rest are not. Suppose that the project manager randomly selects 2 workers (without replacement). What is the probability that NEITHER of the workers selected are left-handed? Round your answer to two decimal places.

Answer 1

The probability that neither of the workers selected are left-handed is 0.9856.

• Number of workers = 50
• Number of left handed people = 6
• Number of right handed people = 50 - 6 = 44

Therefore, the probability that neither of the workers selected are left-handed will be:

= 1 - (6/50 × 6/50)

= 1 - (0.12 × 0.12)

= 1 - 0.0144.

= 0.9856

In conclusion, the correct option is 0.9856.

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Answer 2

The probability that neither of the workers selected are left-handed is about .77 - got it right on kahn