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Problem of the Week 9/6 and Solution

Problem Set # 1 - Back to School - Due Tues Sept 14th on graph paper to be handed in

09/08/10

1. Ms. Johnson is selecting students to solve last night’s homework problems on the board. The homework assignment has a total of 24 problems and there are 25 students in your class. Ms. Johnson has already chosen students to solve the first fourteen problems and you have not been selected yet. Once selected, a student will not be selected to solve another problem from this assignment. What is the probability that you will not be selected to solve any of the remaining problems?
Express your answer as a common fraction.
Since there are 25 students in your class and 14 students have already been chosen to solve
problems, that leaves 25−14 = 11 students from which to choose for the remainder of the problems.
We are told that the assignment consists of 24 problems which means there are 24−14 = 10 problems left to be solved. Consider the probability that you will be selected to solve one of the remaining ten problems. There are ten ways to achieve a favorable outcome out of a total of eleven possible outcomes. So, P(chosen) = 10/11. Since P(chosen) + P(not chosen) = 1, the probability that you will not be chosen to solve any of the remaining problems is 1/11.

2. Today is the first day of school, and you were so excited that you forgot to pack your lunch to bring to school. Now you must purchase lunch from the cafeteria. Fortunately, you have just enough money to buy one item from each of the following categories: sandwiches, side items, beverages and snacks. Your choices are shown below.

Sandwiches      Side Items                  Beverages              Snacks
Hamburger        French Fries              Fruit Juice            Cherry Pie
Hot Dog            Rice Pilaf                  Water                   Cookie
Tuna Fish          Steamed Veggies      Lemonade            Nachos
Salad                Tea                           Apple                   Banana

How many different lunch combinations can you possibly choose?
One way to solve a problem like this is by using a tree diagram to list the possible combinations.
But that would be very time consuming for this particular problem. So instead we make use of the

Fundamental Counting Principle. It states that if you have an event that can occur m different ways, and another event that can occur n different ways, then these events can occur together m · n different ways. In this case, you may choose from three different sandwiches, four different side items, four different beverages,and five different snacks. Therefore, you can choose 3 · 4 · 4 · 5 = 240 different lunch combinations.