A cafeteria has 5 turkey sandwiches, 6 cheese sandwiches, and 4 tuna sandwiches. There are two students in line and each will take a sandwich. What isthe the probability that the first student takes a cheese sandwich and the next student takes a turkey sandwich?


the answers I hqv4 to choose from are
1/7
2/15
1/14
2/21

Prob(cheese, then turkey)

= (6/15)(5/14)=
= 30/210
= 1/7

(6/15)(5/14) = 2/14 = 1/7

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To find the probability that the first student takes a cheese sandwich and the next student takes a turkey sandwich, we need to calculate the probability of each event happening separately and then multiply them together.

First, let's find the probability that the first student takes a cheese sandwich. There are a total of 5 + 6 + 4 = 15 sandwiches in the cafeteria. Since there are 6 cheese sandwiches, the probability of the first student taking a cheese sandwich is 6/15.

Next, let's find the probability that the second student takes a turkey sandwich. After the first student has taken a sandwich, there will be 15 - 1 = 14 sandwiches remaining in the cafeteria. Since there are 5 turkey sandwiches, the probability of the second student taking a turkey sandwich is 5/14.

Now, multiply the two probabilities together:

(6/15) * (5/14) = 30/210 = 1/7

Therefore, the probability that the first student takes a cheese sandwich and the next student takes a turkey sandwich is 1/7.

So, the correct answer is 1/7.

well, first find out how many cheese and turkey sandwiches there are. so, we add 5+6, which would then equal 11. the leftover 4 we subtract from 11. so the denominator would be 7. then, we divide 1/7 by 2. we do that cause there r 2 kids. the answer would be 1/14.

1/14
is your answer!!!

Please direct your questions to your teacher, Mrs. Pratte!