Asked by john
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?
(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?
Answers
Answered by
Reiny
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
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
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