Question

A shipment of 18 televisions sets contains 3 defective sets. A hotel purchases 8 of these televisions sets. What is the probability that the hotel receives at least one of the defective sets?

Answer 1

Answer:

85.29%

Step-by-step explanation:

We know 18 televisions sets contains 3 defective sets => 15 TV sets are ok

So the total number of possible ways for the hotel to buy = $C^{18} _{8} = 43758$

At least one of the defective sets, so we need to find the probability for 1 defective sets, 2 defective sets and 3 defective sets.

1. The number of ways  the hotel purchases 8 of these televisions sets but have 1 defective one:

P(1) = $C^{15} _{7}$ * $C^{3} _{1}$ = 6435*3 = 19305

2.  The number of ways  the hotel purchases 8 of these televisions sets but have 2 defective one:

P(2) = $C^{15} _{6}$ * $C^{3} _{2}$ = 5005*3= 15015

3.  The number of ways  the hotel purchases 8 of these televisions sets but have 3 defective one:

P(3) = $C^{15} _{5}$ * $C^{3} _{3}$ = 3003*1 = 3003

So the probability that the hotel receives at least one of the defective sets =

(P(1) + P(2) + P(3) ) / P = (19305 +15015 + 3003 ) / 43785*100% = 85.29%