Answer:
Probability distribution for x:
[tex]P(x=0)=0.3644\\\\P(x=1)=0.4373\\\\P(x=2)=0.1749\\\\P(x=3)=0.0233\\\\[/tex]
Step-by-step explanation:
We can model the number of defective sets in the group of TV sets (variable x) as a binomial variable, with sample size=3 and probability of success p=2/7≈0.2857.
The probability of k defective sets in the group is:
[tex]P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}[/tex]
So, we have this probabilty distribution for x:
[tex]P(x=0) = \dbinom{3}{0} p^{0}q^{3}=1*1*0.3644=0.3644\\\\\\P(x=1) = \dbinom{3}{1} p^{1}q^{2}=3*0.2857*0.5102=0.4373\\\\\\P(x=2) = \dbinom{3}{2} p^{2}q^{1}=3*0.0816*0.7143=0.1749\\\\\\P(x=3) = \dbinom{3}{3} p^{3}q^{0}=1*0.0233*1=0.0233\\\\\\[/tex]
