Question

Dana buys a bag of cookies that contains 6 chocolate chip cookies, 9 peanut butter cookies, 5 sugar cookies and 8 oatmeal raisin cookies. What is the probability that Dana randomly selects an oatmeal raisin cookie from the bag, eats it, then randomly selects another oatmeal raisin cookie?

Answer 1

Answer:

$P=\frac{2}{27} \approx 0.074 \approx 7.4\%$

Step-by-step explanation:

In order to solve this problem, we need to start by finding the probability of picking an oatmeal raisin cookies. Probabilities are found by using the following formula:

$Probability=\frac{\#of\,desired\,events}{\# of\,possible\,events}$

In this case, our number of desired events is the number of oatmeal raisin cookies we have available: 8.

Our number of possible events is the total number of cookies in the bag, which is:

6+9+5+8=28

so the probability is:

$P=\frac{8}{28}$

which simplifies to:

$P=\frac{2}{7}$

Next, we need to find the probability of picking one oatmeal raisin cookie again. This time we only have 7 oatmeal raising cookies and 27 cookies left in the bag, so our probability will now be:

$P=\frac{7}{27}$

So, the probability of picking an oatmeal raisin cookie at first and picking one of the same kind again is found by multiplying both probabilities.

$P=\frac{2}{7}*\frac{7}{27}=\frac{2}{27}$

So the probability that Dana randomly selects an oatmeal raisin cookie from the bag, eats it, then randomly selects another oatmeal raisin cookie is:

$P=\frac{2}{27} \approx 0.074 \approx 7.4\%$