Respuesta :
The correct answer is:
20/27 = 0.7407.
Explanation:
To find the probability that the restaurant buys at least 2 non-defective units, we find the probability that they purchase either 2 or 3 non-defective units:
[tex]P(r) = _nC_r(p)^r(1-p)^{n-r} [/tex]
They purchase 3 units total, so this is n.
We want the probability of either 2 or 3 non-defective units; this means r is 2 and r is 3.
In this case we want the probability that the units are non-defective. Since there are 4 defective units, this means there are 12-4=8 non-defective units; this makes the probability of a non-defective unit 8/12 = 2/3. This is p.
Using these, we have:
[tex]P(X=2 or X=3)=_3C_2(\frac{2}{3})^2(1-\frac{2}{3})^{3-2}+_3C_3(\frac{2}{3})^3(1-\frac{2}{3})^{3-3} \\ \\=\frac{3!}{2!1!}(\frac{2}{3})^2(\frac{1}{3})^1+\frac{3!}{3!0!}(\frac{2}{3})^3(\frac{2}{3})^0 \\ \\=3(\frac{4}{9})(\frac{1}{3})+1(\frac{8}{27})(1) \\ \\=\frac{12}{27}+\frac{8}{27}=\frac{20}{27}=0.7407[/tex]
20/27 = 0.7407.
Explanation:
To find the probability that the restaurant buys at least 2 non-defective units, we find the probability that they purchase either 2 or 3 non-defective units:
[tex]P(r) = _nC_r(p)^r(1-p)^{n-r} [/tex]
They purchase 3 units total, so this is n.
We want the probability of either 2 or 3 non-defective units; this means r is 2 and r is 3.
In this case we want the probability that the units are non-defective. Since there are 4 defective units, this means there are 12-4=8 non-defective units; this makes the probability of a non-defective unit 8/12 = 2/3. This is p.
Using these, we have:
[tex]P(X=2 or X=3)=_3C_2(\frac{2}{3})^2(1-\frac{2}{3})^{3-2}+_3C_3(\frac{2}{3})^3(1-\frac{2}{3})^{3-3} \\ \\=\frac{3!}{2!1!}(\frac{2}{3})^2(\frac{1}{3})^1+\frac{3!}{3!0!}(\frac{2}{3})^3(\frac{2}{3})^0 \\ \\=3(\frac{4}{9})(\frac{1}{3})+1(\frac{8}{27})(1) \\ \\=\frac{12}{27}+\frac{8}{27}=\frac{20}{27}=0.7407[/tex]
