Question

A television store owner figures that 35 percent of the customers entering his store will purchase and ordinary television set; 25 percent will purchase a color 4) IS store on a certain day, what is the probability that 8 customers will purchase ordinary sets, 5 customers will purchase color sets and 2 will purchase nothing?

Answer 1

This is a problem involving a multinomial distribution.
Given,

• n = number of customers = 15
• n1 = number of customers purchasing ordinary sets = 8
• n2 = number of customers purchasing color sets = 5
• n3 = number of customers purchasing nothing = 2
• p1 = probability of a customer purchasing an ordinary set = 0.35
• p2 = probability of a customer purchasing a color set = 0.25
• p3 = probability of a customer purchasing nothing = 0.40
  

Now, to find the probability that out of 15 customers, 8 customers will purchase ordinary sets, 5 will purchase color sets and 2 will purchase nothing. Let it be denoted by P.

Therefore, $P = \frac{15!}{8!\; 5!\; 2!} (0.35)^8 (0.25)^5 (0.40)^2 = 0.0048$ .