Question

A shipment of 8 TV sets includes 3 that are defective. If 4 of the sets are chosen at random.

A)Find the probability of having no defective sets.

B.Find the probability of having three defective sets.

Answer 1

Solution

Given that ,

p = 3 / 8 = 0.375

n = 4

Using binomial probability formula ,

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x}

A)

P(X = 0) = (₄ C ₀) * 0.375⁰ * (0.625)⁴

= 0.1526

P(no defective sets) = 0.1526

B)

P(X = 3) = (₄ C ₃) * 0.375³ * (0.625)¹

= 0.1318

P( three defective sets) = 0.1318