8 workers are employed in a factory and of them are excellent in effic

1.	8 workers are employed in a factory and of them are excellent in efficiency where as the rest of them are moderate in efficiency. 2 workers are randomly selected from these 8 workers. Find the probability that(1) both the workers have excellent efficiency(2) both the workers have moderate efficiency(3) one worker is excellent and one worker is moderate in efficiency.
Answer» From 8 workers 3 workers are excellent in efficiency. So the remaining 5 workers are moderate in efficiency. Total number of primary outcomes of selecting 2 workers at random from 8 workers is n = ⁸C₂ =\( \frac{8×7}{2×1}\) = 28 (1) A = Event that the selected both the workers have excellent efficiency ∴ Favourable outcomes for the event A is m = ³C₂ × ⁵C₂ = 3 × 1 = 3. Hence, P(A) = \(\frac{m}{n} = \frac{3}{28}\) (2) B = Event that the selected both the workers have moderate efficiency ∴ Favourable outcomes for the event B is m = ⁵C₂ × ³C₀ = 10 × 1 = 10. Hence, P(B) =\( \frac{m}{n} = \frac{10}{28} = \frac{5}{14}\) (3) C = Event that in the selected two workers one worker is excellent and ope worker is moderate in efficiency. ∴ Favourable outcomes for the event C is m = ³C₁ × ⁵C₁ = 3 × 5 = 15. Hence. P(C) = \(\frac{m}{n }= \frac{15}{28}\)

8 workers are employed in a factory and of them are excellent in efficiency where as the rest of them are moderate in efficiency. 2 workers are randomly selected from these 8 workers. Find the probability that(1) both the workers have excellent efficiency(2) both the workers have moderate efficiency(3) one worker is excellent and one worker is moderate in efficiency.

Answer»

From 8 workers 3 workers are excellent in efficiency. So the remaining 5 workers are moderate in efficiency. Total number of primary outcomes of selecting 2 workers at random from 8 workers is

n = ⁸C₂ =\( \frac{8×7}{2×1}\) = 28

(1) A = Event that the selected both the workers have excellent efficiency

∴ Favourable outcomes for the event A is m = ³C₂ × ⁵C₂ = 3 × 1 = 3.

Hence, P(A) = \(\frac{m}{n} = \frac{3}{28}\)

(2) B = Event that the selected both the workers have moderate efficiency

∴ Favourable outcomes for the event B is m = ⁵C₂ × ³C₀ = 10 × 1 = 10.

Hence, P(B) =\( \frac{m}{n} = \frac{10}{28} = \frac{5}{14}\)

(3) C = Event that in the selected two workers one worker is excellent and ope worker is moderate in efficiency.

∴ Favourable outcomes for the event C is

m = ³C₁ × ⁵C₁ = 3 × 5 = 15.

Hence. P(C) = \(\frac{m}{n }= \frac{15}{28}\)

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