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Probabilities using Counting Techniques

Probabilities using Counting Techniques. In order to compute probability, we need to have a sample space and the number of times that the event happens. P(A) = n(A) n(S).

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Probabilities using Counting Techniques

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  1. Probabilities using Counting Techniques

  2. In order to compute probability, we need to have a sample space and the number of times that the event happens. P(A) = n(A) n(S) This principle still applies to this section, although now we are talking about situations in which we need to calculate a combination.

  3. Example. Two brothers are playing on a badminton team with five other people. What are the chances that the two brothers will play together with the older one on the left side and the younger one on the right side of a doubles match. Ask yourself… does this combination care about the order?

  4. Therefore to tell us the sample space, we need to find out the number of combinations using the permutations formula. N(S) = 7P2 = 42 The specific order of the brothers in the right side and the left side can only happen in one way. P(A) = 1/42 What if the order doesn’t matter?? Combinations! nCr

  5. A focus group of 4 people needs to be created from 6 students and 7 teachers. What is the probability of creating a 4-member committee not comprised entirely of students? In this situation, we need to figure out what the probability is to choose a group from entirely students. Once we do this we can use 1 – P(A) to get P(A’).

  6. Probability using the fundamental counting principle. What is the probability that two or more students out of a class of 24 will have the same birthday? Assume that no students were born on February 29. The best way to approach this is from the other side (complimentary). Therefore we want to find out the prob that two or more students will not have the same birthday.

  7. We know the first student will have a 365/365 chance to have his birthday with no similarities. Since the second student can’t have the same birthday, his chance is 364/365. The next student will be 363/365. This will continue for the class of 24. P(A’) = n(A’) / n(s) = 365/365 x 364 / 365 x…x 342/365 = 0.462 P(A) = 1 – P(A’) = 0.538

  8. Homework Pg 324 # 1, 3, 5, 6, 8, 9, 10

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