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 = [tex]C^{18} _{8} = 43758[/tex]
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) = [tex]C^{15} _{7}[/tex] * [tex]C^{3} _{1}[/tex] = 6435*3 = 19305
2. The number of ways the hotel purchases 8 of these televisions sets but have 2 defective one:
P(2) = [tex]C^{15} _{6}[/tex] * [tex]C^{3} _{2}[/tex] = 5005*3= 15015
3. The number of ways the hotel purchases 8 of these televisions sets but have 3 defective one:
P(3) = [tex]C^{15} _{5}[/tex] * [tex]C^{3} _{3}[/tex] = 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%
