Class lecture 4-5 and 4-6 PDF

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4–5 Multiplication Rule: Complements and Conditional Probability Complements: Understand the meaning of “at least one” : One or more event The complement of “one or more” is : None We know that probability of occurrences of the entire sample space is 1 P (at least one) = 1 – P (none) [The reason we are using this concept is to reduce work. P(one or more) needs adding all the probabilities for one or more. Instead, we can just get one probability (of the complement) and subtract it from 1. ] ex) If we toss a coin three times, what is the probability that we would have at least one head? (one or more) Let event A = At least one head ( one or more) Then = no heads (all tails) P( ) = P ( tail and tail and tail)

=

=

P (A) = 1 – P ( ) = 1 -

=

Conditional Probability: P(B|A)

*Formal Multiplication Rule P(A and B) = P(A) P(B|A) from here, we can obtain P(A and B) / P(A) = P(B|A)

ex) Drug testing Did the Subject Actually Use Marijuana? Yes No ____________________________________________________________________ Positive test result 119 24 (Test indicated that (true positive) (false positive) marijuana is present) Negative test result (Test indicated that Marijuana is absent)

3 (false negative)

154 (true negative)

a. Find the probability that a randomly selected person had a positive test result given that the person actually did not use marijuana? 24/ 178 = .1348

Another way by the multiplication rule:

= 24/178 = .1348

b. Find the probability that a randomly selected person had a positive test result given that the person actually used marijuana.

119/122 = .9754

Another way :

= 119/122 = .9754

Interpretation: a. Among the non-marijuana users, the probability that the person would be test

positive is .1348 b. Among the marijuana users, the probability that the person would be test positive is .9754

More examples 1. Find the probability that a couple will have all four girls.

2. Probability of the complement of question no.1: “all four girls” ( means “no boys” ) So the complement of no boys is “at least one or more boys”. P(at least one or more boys) is 1 – (no boys) =

3. Find the probability that a couple will have all 4 children of the same gender.

4. Find the probability that when 4 digits are randomly selected (from 0 to 6 inclusive) with replacement, that we will get at least one 5. P(at least one 5) = 1- P(no 5)

5. Birth in China: The probability of a baby being a boy is .545. Many couples are allowed to have only one child. Among the next five randomly selected births in china, what is the probability that at least one of them is a girl. P (at least one girl) = 1 P(No girls) = 1 – [P(boys)]5 1 ( .545 )5 = 6. You have an alarm clock that has a failing rate of 4%. a. What is the prob. that it does not fail :

b. What if you have 3 alarm clocks, what is the prob. that not all of three fail?

4 - 6 Counting Rule (1) Fundamental Counting Rule In a sequence of three events, if the first event can occur m ways, the second event , n ways, and the third event can occur k ways, the total number of outcomes is : mxnxk ex) If you have two hats, four shirts, and three pairs of pants, how many distinct ways are there to coordinate your look?

ex) What is the number of the possible combinations of a social security number? (How many digits are there? How many numbers are there to choose from? Is it with or without replacement when selecting a number?)

[ so if someone says s/he got your social security number by random, is it likely to happen? ]

ex) You are trying to get an industrial license, and you’re required to pass 7 tests. What is the total number of the possible outcomes of the 7 tests? -

Each test has a result of pass or fail, so there are

(2) Factorial Rule n! = n x (n-1) x (n-2) x - - - x 2 x 1 4! = 4 x 3 x 2 x 1 = 24 0! = 1

ex) Assume that you have five chores to do (A, B, C, D, E). How many ways are there to do all your chores? Would it be with replacement or without replacement? [ you don’t do the same chores twice]

Winning a lottery. Which one is more difficult to win? ex 1) matching 5 numbers out of 39 in any order. ex 2) matching 5 numbers out of 39, but the order (the sequence) also needs to match the way they were drawn. (Why is it more difficult? The probability is smaller because there are more possible combinations of sequences for the second one. ) (For ex 1), if the winning numbers are (1, 3, 5, 7, 9), any combination like (5, 3, 1, 7, 9) and (7, 1, 3, 5, 9) are all treated as same as (1, 3, 5, 7, 9), so these numbers are considered as just one outcome. But for ex 2), (1, 3, 5, 7, 9) is one outcome. (5, 1, 3, 7, 9), is one outcome, and so forth. So more outcomes are possible for ex2). Consequently, it is more difficult to win. ) To compute For example 1 (the order does not count.), We use combination rule. (3) Combination Rule (the order does not count): If there are n different items available, and we select r out of n without replacement, the number of combinations is: To do ex 1) nCr=

39

C5 =

=

=

Ti -83/84 plus functions for permutations or combinations: 39

C5

Press 39 - Press [Math] – Highlight [PRB] Choose – [nCr] - Press [Enter] - Press 5 – [Enter]

To compute example 2 (when the order matters), we use permutation rule. (4) Permutation Rule 1 (the order matters) If there are n different items available, and we select r out of n without replacement, the number of permutations is: nPr=

the result of ex 2) is:

39

P 5=

=

=

(5) Permutation Rule 2 (the order matters) If there are some identical items, we use the following formula: (for n1 alike, n2 alike, and so forth) ex) How many ways can the word, Mississippi be arranged? Statistics be arranged?

More Counting Examples 1. When randomly guessing the first and second winners of a competition out of 16, how many ways are there? Permutation or Combination rule?

2. The winning numbers for the current California Daily 4 lottery are 5, 0, 0, and 4 in that exact order. Because order counts, do calculations for this lottery involve either of the two permutation rules presented in this section? Why or why not? If not, what rule does apply?

3. If you are about to donate your 7 clothing items one by one, how many orders are there?

4. There is a list of 6 presidents (Carter, Reagan, Nixon, Ford, J. F Kennedy, and Johnson) a) How many ways are there to list the 6 presidents?

What is the probability of choosing the correct one (randomly)?

b) (Another way of computing the probability) What is the probability of getting the list of the 6 presidents in the correct chronological order ?

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When you want to figure out the number of ways or choices from the total number of items (without replacement): Use Factorial

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When you want to figure out the number of ways of choices by choosing some items from the total (without replacement) : Use Combination or Permutation...