Question

In an industry, the workers are having 25% chance of suffering from an occupational disease. Among 6 randomly selected workers, what is the probability that at most 2 will contract the disease?

Answer 1

Let x be the number of workers contracting disease is a Binomial variable with parameters n = 6, 

p = p [a worker contract the disease] 

= 25% = 0.25.

∴ q = 0.75.

The p.m.f is –

P(x) = nc_xp^xq^{n – x}; x = 0,1,2, …… n

= 6c_x 0.25^x (0.75)^6-x; x = 0,1,2, …….. 6

p (at most 2 workers contract the disease)

= p(x ≤2)

= (x – 0) + p(x = 1) + p(x = 2)

= 6c₀ (0.25)⁰ (0.75)⁶ – 0+ 6c₁ (0.25)¹ (0.75)^{6 – 1} 

= 6c² (0.25)² (0.75)^{6 – 2}

= 1 ×1 × 0.1799 + 6 × 0.25 × 0.2373 + 15 × 0.0625 × 0.3164

= 0.1779 + 0.3559 + 0.2966

= 0.8304.