Assignment of Probabilities

Example 2| Example 4| Example6| Combination or Permuation | Useful Web Resources| Solutions

Sample Spaces and Events

 

 

Multiplication Rule

If a procedure A can be done in m ways and, for each way of doing procedure A, procedure B can be done in n ways, then A followed by B can be done in m x n ways.

     

    Example 1:

      Consider an example that involves choosing toppings for a pizza. There are 5 choices for the type of meat, 4 choices for the types of vegetables, 1 choice for the type of fruit (pineapple), and 4 choices for the type of cheese. How many different pizzas can you make if you must put one type of meat, vegetable, fruit and cheese on your pizza.?

      Solution:

      For any meat you choose, there are 4 vegetables to choose from: So there are (5)(4) ways. For any vegetable you choose, there is one fruit to choose from: There are (5)(4)(1) ways to do this. Then you have 4 types of cheeses to choose from: (5)(4)(1)(4)=80. So there are 80 different pizzas you can make from these toppings.

       

    Example 2:

    Consider tossing a fair coin 8 times. How many samples are in the sample space?

    1. 2x8
    2. (8)(8)
    3. (2)(2)(2)(2)(2)(2)(2)(2)=2^8

    Solution

Definition 1.4-1: Permutation

An ordered arrangement of r items selected from n distinct items is called a permuation of n distinct items taken r at a time.

Example 3:

    How many ways can you arrange 5 of 10 choir members on a stage?

     

    Solution:

    We can apply the multiplication rule here. There are 10 choices for the first position, 9 for the second, 8 for the third etc. So we have (10)(9)(8)(7)(6)= 10!/(10-5)! = 10!/5! = 10P5 =30,240

    Example 4:

    Suppose there are five research projects that need to be completed, each requiring one scientist. In how many different ways can five scientists from a group of 100 be assigned to these 5 research projects?

    1. 100! / 95!
    2. 100! / 5!
    3. (100)(100)(100)(100)(100)=100^5

Solution

Definition 1.4-2: Combination

An unordered arrangement of r items selected from n distinct items is called a combination.

 

Example 5:

How many ways can you choose a group of 5 students from 10 to be in the choir?

Solution:

You can arrange 5 students from 10 in 10P5 ways. But since order does not matter, we must divide by 5! to get rid of the over counting. So we get (10!) / (5!)(5!) = 10C5.

 

    Example 6:

    How many ways can we arrange the letters in STATISTICAL?

    1. 11! / (3!)(2!)(2!)(2!)
    2. 11!
    3. 11! / (3)(2)(2)(2)

    Solution

Permutation or Combination?

Please enter your answer in the form nCr,nPr,or as an integer and press ENTER.

    A: Fred is a casting agent in charge of hiring three extras for a movie. (These extras will be members of a crowd in the background and will have no speaking parts.) So far he has had 32 people audition.

      Combination
      Permutation

    In how many ways could this hiring job be done?


    B: How many ways can Karen arrange her six antique candlesticks on her mantle?

      Combination
      Permutation
    C: Eight students each submit an essay for competition. In how many ways can 3 essays be chosen to receive a certificate of merit?

    Combination
    Permutation

    D: A Hospital ward contains 12 patients, 8 of whom have heart disease. If an intern randomly selects 4 of the patients to examine, what is the probability that 2 of them will have heart disease and 2 of them won't?

      Combination
      Permutation

      Please enter your answer in a fraction form. eg 12/43

    Solution

    These questions were taken from http://homepages.ius.edu/pwills/T102/reviews/7-5review.pdf

     

     

    Useful Web Resources

    Stats: Probability Rules

    Multiplication Rule

    Permutations and Combinations

    Counting: Permutations and Combinations

     

    Solutions

    Solution to Example 2

    The answer is C because for each of the 8 flips, there are 2 possible outcom -either head or tail. So the sample space will contain 2^8 points.

    Solution to Example 4

    You have to choose r = 5 scientists for the 5 research projects from a batch of n=100.

    So nPr= 100!/(100-5)!= 100!/95! = 9,034,502,400

    Solution to Example 6

    There are 3 T's, 2 S's, 2 I's and 2 A's in STATISTICAL. So there are 11! ways to arrange the 11 letters, but we must divide by (3!)(2!)(2!)(2!) so that we don't over count the repeating T's,S's I's and A's. So the answer is 11!/(3!)(2!)(2!)(2!)

    Solution to "Permutation or Combination?

    A: This question is a combination problem, because we are not concerned with the order that the three extras are hired in. For example choosing person A, then person B then person C is the same as hiring person B first, then person A, and then person C. The two cases represent the same three people being hired. So there are 32C3=4960 ways to hire 3 people from the 32 people.

    B: The answer to this question is 6!. This can be expressed as a permutation because 6P6 = 6!/(6-6)!=6!/0!= 6!/1= 6!

    C: This is question is a combination problem because the order in which you give out the certificates of merits do not matter. So we there are 8C3= 56 ways that this can be done.

    D: Let A be the number of ways to choose 4 patients to examine, where two of them have heart disease, and two of them don't. Let B be the number of ways to choose any 4 patients . So from the 8 patients that have heart disease, we want to choose 2 of them. From the 12-8=4 patients that don't have heart disease we want to choose 2 of them. So A=(8C2)(4C2)=168. The number of ways to choose 4 patients from 12 without any restrictions is 12C4=495. So our final answer is A/B= 168/495

     

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