Respuesta :
Solution
The total number nof cookies in the bag is:
[tex]4+6+7+7=24\text{ cookies}[/tex]- Thus, we can write the probabilities of choosing any of the cookies are given as:
[tex]\begin{gathered} P(\text{chocolate chip)}=\frac{4}{24} \\ \\ P(\text{Peanut Butter)}=\frac{6}{24} \\ \\ P(\text{Sugar Cookies)}=\frac{7}{24} \\ \\ P(\text{Oatmeal)}=\frac{7}{24} \end{gathered}[/tex]- Now, let us analyze Trey's random choices
Choice 1:
He chose a sugar cookie at first.
- Thus, the probability of this choice is
[tex]P(\text{Sugar Cookies})=\frac{7}{24}[/tex]Choice 2:
- He chose an oatmeal cookie in the second choice.
- But he already had one cookie before this choice, thus, the total number of cookies is one less. That is, 23 not 24.
- Thus, the probability of choosing the oatmeal cookie is given as:
[tex]P(\text{Oatmeal)}=\frac{7}{23}[/tex]- Because these choices do not interfere with one another (i.e. they are mutually exclusive), we can apply the AND probability formula to calculate the probability that Treys chose a Sugar Cookie first, and then an Oatmeal cookie next.
- The AND probability is given as
[tex]P(A\text{ AND }B)=P(A)\times P(B)[/tex]- Thus, we can find the probability that Trey chooses Sugar Cookies first and Oatmeal Cookie next as follows:
[tex]\begin{gathered} P(\text{Sugar Cookies AND Oatmeal)}=\frac{7}{24}\times\frac{7}{23} \\ \\ P(\text{Sugar Cookies AND Oatmeal)}==\frac{49}{552} \end{gathered}[/tex]Final Answer
The answer is
[tex]P(\text{Sugar Cookies AND Oatmeal)}==\frac{49}{552}[/tex]