Question

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 1

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) = \(\frac{m}{n }= \frac{15}{28}\)