Phyllis made some sandwiches for a family reunion.She made 12 turkey sandwiches, 10 ham sandwiches, and 8 cheese sandwiches . She wrapped the sandwiches in paper but forgot to label them.

(Part A) The first person took 2 candwiches what is the probability that the first sandwich was turkey and the second ham?

(Part B) Suppose the first two sandwiches taken were actually both ham. What is the probability that the third sandwich taken is cheese?

Very similar to your previous question

A
prob (turkey then ham) = (12/30)(10/29) = ...

B
Two hams are gone, so 28 sandwiches left
prob(next one is cheese) = 8/28

To solve these probability problems, we first need to determine the total number of possible outcomes and the number of favorable outcomes.

Let's calculate the total number of possible outcomes (the total number of ways the sandwiches can be chosen) for both parts.

(Part A)
Phyllis made 12 turkey sandwiches, 10 ham sandwiches, and 8 cheese sandwiches. Hence the total number of possible outcomes is the total number of sandwiches made:
Total number of possible outcomes = 12 (turkey sandwiches) + 10 (ham sandwiches) + 8 (cheese sandwiches) = 30

(Part B)
The same total number of possible outcomes applies here as well; hence the total number of possible outcomes also equals 30.

Now let's calculate the favorable outcomes for each part.

(Part A)
For the first sandwich to be turkey and the second sandwich to be ham, we have to choose 1 out of the 12 turkey sandwiches and 1 out of the 10 ham sandwiches.
Hence, the total number of favorable outcomes for this part is:
Total number of favorable outcomes = 12 (turkey sandwiches) * 10 (ham sandwiches) = 120

(Part B)
Since the first two sandwiches were ham, we have already taken them out of the total possible outcomes, leaving 28 sandwiches. Out of these 28 sandwiches, there are 8 cheese sandwiches.
Hence, the total number of favorable outcomes for this part is:
Total number of favorable outcomes = 8 (cheese sandwiches)

Now we can calculate the probabilities for each part.

(Part A)
Probability of choosing a turkey sandwich first:
P(turkey first) = Number of favorable outcomes / Total number of possible outcomes
P(turkey first) = 12 / 30 = 2 / 5

Probability of choosing a ham sandwich second (given that the first sandwich was turkey):
P(ham second | turkey first) = Number of favorable outcomes / Total number of possible outcomes
P(ham second | turkey first) = 10 / 29

To find the probability of both events occurring (turkey first and ham second), we multiply the probabilities:
P(turkey first and ham second) = P(turkey first) * P(ham second | turkey first)
P(turkey first and ham second) = (2 / 5) * (10 / 29)
P(turkey first and ham second) = 20 / 145 ≈ 0.1379

Therefore, the probability that the first sandwich was turkey and the second sandwich was ham is approximately 0.1379 or 13.79%.

(Part B)
Probability of choosing a cheese sandwich third (given that the first two sandwiches were both ham):
P(cheese third | ham first and ham second) = Number of favorable outcomes / Total number of possible outcomes
P(cheese third | ham first and ham second) = 8 / 28 = 2 / 7 ≈ 0.2857

Therefore, the probability that the third sandwich taken is cheese, given that the first two sandwiches were both ham, is approximately 0.2857 or 28.57%.