Respuesta :
Answer:
0.0476
Step-by-step explanation:
Given:
Armando buys a bag of cookies that contains 6 chocolate chip cookies, 7 peanut butter cookies, 4 sugar cookies and 7 oatmeal cookies.
Question asked:
What is the probability that he chosen randomly 2 sugar cookies from the bag?
Solution:
As we know:
[tex]Probability =\frac{Favourable \ outcome}{Total\ outcome}[/tex]
First of all we will find favorable outcome for choosing 2 sugar cookies from the bag randomly:
Favorable outcome for choosing 2 sugar cookies out of 6 chocolate chip cookies = [tex]^{6} C_{0}[/tex]
Favorable outcome for choosing 2 sugar cookies out of 7 peanut butter cookies = [tex]^{7} C_{0}[/tex]
Favorable outcome for choosing 2 sugar cookies out of 4 sugar cookies = [tex]^{4} C_{2}[/tex]
Favorable outcome for choosing 2 sugar cookies out of 7 oatmeal cookies = [tex]^{7} C_{0}[/tex]
Thus, total Favorable outcome for choosing 2 sugar cookies =
By using: [tex]^{n} C_{r}=\frac{n!}{(n-r)!\ r!}[/tex]
[tex]^{6} C_{0}\times^{7} C_{0}\times^{4} C_{2}\times^{7} C_{0}[/tex]
[tex]\frac{6!}{(6-0)!\ 0!} \times\frac{7!}{(7-0)!\ 0!}\times\frac{4!}{(4-2)!\ 2!}\times\frac{7!}{(7-0)! \ 0!}[/tex]
[tex]1\times1\times\frac{4\times3\times2}{2\times2} \times1=\frac{24}{4} =6[/tex]
Total outcome for choosing 2 sugar cookies out of 24 cookies =
[tex]^{24} C_{2}[/tex] = [tex]\frac{24!}{(24-2!\ 2!)}[/tex]=[tex]\frac{24!}{(24-2)!\ 2!} =\frac{24\times23\times22!}{22!\ 2!} =\frac{552}{2}[/tex]
Now,
[tex]Probability =\frac{Favourable \ outcome}{Total\ outcome}[/tex]
[tex]=\frac{6}{\frac{552}{2} }\\ \\ = 6\div\frac{552}{2}[/tex]
[tex]=6\times\frac{2}{552} \\\\ =\frac{12}{252} \\\\ =0.0476[/tex]
Thus, the probability that Armando reaches in the bag and randomly selects 2 sugar cookies from the bag is 0.0476
