Question

In an office 20% of the employees are females. In a random sample of 5 employees obtain the probability that there are – (a) no females (b) females only.

Answer 1

Let x be the no. of the female employees is a Binomial variate with parameter n = 5. 

p = p [an employee is a female] 

= 20% = 0.2. 

∴ q = 1 – p = 0.8

The p.m.f. is –

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

= 5 c_x (0.2)^x (0.8)^{5 – x} ; x = 0,1,2, … 5

(a) p (no females) = p(x = 0)

= 5c₀ (0.2)⁰ (0.8)^{5 – x}

= 1 × 1 × 0.3277

= 0.3277

(b) p (females only) 

= p(x = 5) 

= 5C₅ (0.2)⁵ (0.8)^5-5 

= 1 × 0.00032 × 1 

= 0.00032.