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Probability-III (Permutations and Combinations) Probability-III (Permutations and Combinations)

Probability-III (Permutations and Combinations) - PowerPoint Presentation

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Probability-III (Permutations and Combinations) - PPT Presentation

QSCI 381 Lecture 11 Larson and Farber Sect 34 PermutationsI A is an ordered arrangement of objects The number of different arrangements permutations of n different objects is ID: 759741

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Slide1

Probability-III(Permutations and Combinations)

QSCI 381 – Lecture 11

(Larson and Farber, Sect 3.4)

Slide2

Permutations-I

A is an ordered arrangement of objects. The number of different arrangements (permutations) of n different objects is n! (n-factorial).Key note: We are interested in the order of things.

Permutation

Slide3

3! = 3 x 2 x 1 = 6

1 2 3

Permutations-I

Slide4

Permutations-II

The number of permutations of n distinct objects taken r (r <n) at a time is:

Returning to our earlier example; selecting 4 of 10 animals to age.

Here n=10 and r=4 so

Slide5

There are forty-three republican candidates hoping to be in the top three candidates at each primary election. In each state, how many ways can the candidates finish first, second, and third?nPr = 43P3 = 43!/(43-3)! = 43!/40! = 43 x 42 x 41 74,046 primary standing options

Permutations-II

Slide6

Permutations-III

The number of of n objects when n1 are of one type, n2 are of another type, etc. is:

Distinguishable

permutations

= 6!/(3! x 2! X 1!)

= (6 x 5 x 4)/(2 x 1)

= 120/2 = 60

Slide7

Permutations-III(Example)

You have five sockeye, four pink and four chum salmon. How many ways can they be arranged?

Slide8

Permutations-III(Example)

You have five sockeye, four pink and four chum salmon. How many ways can they be arranged?

Slide9

Combinations-I

A is a selection of r objects from a group of n objects without regard to order and is denoted by . The number of combinations of r objects selected from a group of n objects is:Key note: We are not interested in the order of things.

Combination

Slide10

Combinations-II

Example. We have fish from three species (A, B, C) and wish to select two of them. How many ways can we do this:(A,B), (B,C), (A,C). 3C2=3It is not(A,B),(B,C),(A,C),(B,A),(C,B),(C,A) Because order does not count.

3!/(3-2)!2! = 6/2 = 3

Slide11

Combinations-III(Example)

You wish to test the impact of five treatments. However, your experiment can only consider three treatments at once. How many experiments do you need to conduct so that all combinations of the three treatments are examined?

Hint: after the lecture write down all the combinations after naming them A, B, C, D, E.

Slide12

Combinations -IV

The number of combinations of r objects chosen from n distinct objects allowing for repetitions is:

Slide13

Combinations vs Permutations-I

Permutations relate to the number of ways of selecting

r

objects from

n

when the order of the items matters

.

Combinations relate to the number of ways of selecting

r

objects from

n

when the order does not matter

.

Slide14

Combinations vs Permutations-II

Without repetitionWith repetitionPermutationsCombinations

Slide15

Relationship to Probability-I

If there are n people together - what is the probability that all have different birthdays:Total number of permutations of birthdays: 365n (fundamental counting rule 365x365…)Number of ways that n people can have different birthdays: 365Pn=365!/(365-n)!For n=23, the ratio 365P23 / 36523 =0.493.

Permutation without repetition

Slide16

More on Birthdays and Probability

Number of people

Slide17

Relationship to Probability-II

A subgroup of four people is selected at random from a group of 5 married couples. What is the probability that the selection consists of two married couples:Number of ways of selecting 4 people from 10 is: 10!/(6!.4!)Number of ways of selecting 2 couples from 5 couples is : 5!/(3!.2!)The ratio of these is 0.0476.

Combination without repetition

Slide18

Relationship to Probability-III

There are 33 candidates in an election to a Committee of 3. What is the probability that Smith, Jones and Williams are elected (assume that elections are random):

We are interested in one combination of all possible combinations.

There

33

C

3

combinations of 3 selected from 33 so the probability 1/

33

C

3

=0.000183.

Hint. The factorial function in EXCEL is “Fact(x)” and there are also functions “

Combin

(A,B)” and “

Permut

(A,B)”.

Slide19

Relationship to Probability-IV

An experiment involves three counting devices. Each has a probability of failure of 0.02.

What is the probability that one fails?

What is the probability that none fails?

What is the probability at least two don’t fail?

Slide20

Common Mistakes

Assuming probabilities have a memory

: You toss a (fair) coin and get

eight heads

– what was the probability of this and what is the probability of an eight head

?

0.004 0.5 0.25

Adding probabilities incorrectly

. Why is the probability of surveying a pregnant female bowhead whale not 0.7 if the probability of encountering a female is 0.5 and 20% of the population is pregnant?