Question

Trey buys a bag of cookies that contains 4 chocolate chip cookies, 6 peanut butter cookies, 7 sugar cookies and 7 oatmeal cookies.What is the probability that Trey reaches in the bag and randomly selects a sugar cookie from the bag, eats it, then reaches back in the bag and randomly selects an oatmeal cookie? Give your answer as a fraction, or accurate to at least 4 decimal places.

Answer 1

The total number nof cookies in the bag is:

`4+6+7+7=24\text{ cookies}`

- Thus, we can write the probabilities of choosing any of the cookies are given as:

`\begin{gathered} P(\text{chocolate chip)}=(4)/(24) \\ \\ P(\text{Peanut Butter)}=(6)/(24) \\ \\ P(\text{Sugar Cookies)}=(7)/(24) \\ \\ P(\text{Oatmeal)}=(7)/(24) \end{gathered}`

- Now, let us analyze Trey's random choices

Choice 1:

He chose a sugar cookie at first.

- Thus, the probability of this choice is

`P(\text{Sugar Cookies})=(7)/(24)`

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:

`P(\text{Oatmeal)}=(7)/(23)`

- 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

`P(A\text{ AND }B)=P(A)* P(B)`

- Thus, we can find the probability that Trey chooses Sugar Cookies first and Oatmeal Cookie next as follows:

`\begin{gathered} P(\text{Sugar Cookies AND Oatmeal)}=(7)/(24)*(7)/(23) \\ \\ P(\text{Sugar Cookies AND Oatmeal)}==(49)/(552) \end{gathered}`

Final Answer

The answer is

`P(\text{Sugar Cookies AND Oatmeal)}==(49)/(552)`