Chapter 2((probability theory ))



chapter 2((probability theory ))

((1-2))count operation

1-(n factorial) is defined as the product of all the integers from 1 to n (the order of multiplying does not matter) .we can use factorial to determine the number of arrangement of n objects if we select n of them each time.

We write "n factorial" with an exclamation mark as follows: n!

n! = (n)(n − 1)(n − 2)...(3)(2)(1)

2-(permutation) An arrangement (or ordering) of a set of objects is called a permutation. (We can also arrange just part of the set of objects.)In a permutation, the order that we arrange the objects in is important

The number of permutations of n distinct objects taken r at a time, denoted by Pnr,where repetitions are not allowed, is given by

[pic]

3-(permutation with replication) An arrangement of selected (n) from (n) object but with replication in the objects which it is given by:-

p=(n!)/(n1!.n2!…….nk!)

3-( A combination) of n objects taken r at a time is a selection which does not take into account the arrangement of the objects. That is, the order is not important.

And given by

[pic]

Note 1

We can make the ways of count in the following block diagram :-

Note2- if we select (r) objects sequentially from (n) objects with replacement and the order is important then we can find the count as :-

Per(n,1).per(n,1).per(n,1)….. ((r times )

Note3- if we select (r) objects sequentially from (n) objects without replacement and the order is important then we can find the count as :-

Per(n,1).per(n-1,1).per(n-2,1)….. ((r times )

Note4- if we select (r) objects sequentially from (n) objects with replacement and the order is not important then we can find the count as :-

comp(n,1).comp(n,1).comp(n,1)….. ((r times )

Note5- if we select (r) objects sequentially from (n) objects without replacement and the order is not important then we can find the count as :-

comp(n,1).comp(n-1,1).comp(n-2,1)….. ((r times )

(2-2) The Binomial Theorem

We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.

Properties of the Binomial Expansion (a + b)n

• There are n + 1 terms.

• The first term is an and the final term is bn.

• Progressing from the first term to the last, the exponent of a decreases by 1 from term to term while the exponent of b increases by 1. In addition, the sum of the exponents of a and b in each term is n.

• If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.

The binomial expansion (a+b)n can be written as

[pic]

or

[pic]

|[pic] | |

Note

The r th term of the expansion of [pic]is:

[pic]

(3-2)Pascal's Triangle

We note that the coefficients (the numbers in front of each term) follow a pattern. [This was noticed long before Pascal, by the Chinese.]

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

Examples

|1.  Expand  [pic]. |Let a = x, b = 2, n = 5 and substitute.  (Do not substitute a value for k.) |

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    [pic]

[pic]

[pic]

 

 

| Find the 5th term of  [pic]. |Let r = 5, a = (3x), b = (-4), n = 12 and substitute.  |

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[pic]

(4-2) probability

1)) definitions

1-Any experiment whose outcome cannot be predicted in advance, but is one of the set of possible outcomes, is called a random experiment.

2-A sample space is a set of all possible outcomes for an activity or experiment

|Rolling a die |{1, 2, 3, 4, 5, 6} |

|Tossing a coin |{ Heads, Tails} |

|Drawing a card from a standard deck |{52 cards} |

|Drawing one marble from the bottle |{8 marbles} |

|Rolling a pair of dice |{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) |

| |(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) |

| |(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) |

| |(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) |

| |(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) |

| |(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} |

1- An event is the outcome or a combination of outcomes of an experiment. In other words, an event is a subset of the sample space.

2- Probability   If a trial results in n-exhaustive, mutually exclusive and equally likely cases and m of them are favourable to the occurrence of an event A, then the probability of the happening of A, denoted by P(A), is given by:

P(A) = m/n.

2)) properties of probability

We have the following properties of the probability function:

[pic]

The previous properties represent formulas currently used in probability calculus on a finite field of events.

      Property (P9) is the main calculus formula for applications in finite cases.

      In addition, if  {Ω, Σ, P} is a σ-field, we also have the following properties:

[pic]

Independent events. Conditional probability

Let us consider the experiment of tossing two coins and let A – heads on first coin and B – heads on second coin be two events. The occurrence of event A and its probability do not depend on the occurrence of event B, and vice versa. In this case, events A and B are said to be independent (each isindependent of the other).

[pic]

According to this definition, in the previous example we have: P(A and B) = P(A) x P(B) = (1/2) x (1/2) = 1/4.

Consider an urn containing four white balls and three black balls. Two people extract one ball each from the urn. Let A – first person is extracting a white ball and B – second person is extracting a white ball be two events. The probability of event B, in the absence of information about A, is 4/7. If event A has occurred, the probability of event B is 1/2, so event B depends on event A. Therefore, these two events are not independent.

     It is natural to call the probability of event B conditional on event A and to denote it by  P(B│A).

[pic]

     Total probability formula. Bayes’s theorem

[pic]

[pic]

   Bayes’s theorem is a main result in probability theory, which relates the conditional and marginal probability of two aleatory events A and B. In some interpretations of probability, Bayes’s theorem explains how to update or revise beliefs in light of new evidence.

 

|Example: |

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|In tossing a fair die, what is the probability that the outcome is odd or grater than 4? |

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|Suggested answer: |

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|Let E1 be the event that the outcomes are odd. |

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|E1 = {1,3,5} |

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|Let E2 be the event that the outcomes are greater than 4. |

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|E2 = {5,6} |

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|[pic] |

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|[pic] |

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|[pic] |

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|[pic] |

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|[pic] |

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|[pic] |

|Example: |

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|In tossing a die experiment, what is the probability of getting at least 2. |

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|Suggested answer: |

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|Let E be the event that the outcome is at least 2, then |

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|E = {2,3,4,5,6} |

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|EC= {1} |

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|[pic] |

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|[pic] |

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|[pic] |

Below are some solved simple applications:

1) Find the probability of getting a multiple of 2 at a die roll.

 Solution:

     The number of outcomes that are favorable to respective event is three (these are: {2}, {4}, {6}).  The number of equally possible outcomes is six, so the probability is 3/6 = 1/2 = 50%.

 2) There are three pairs of socks of different colors in a basket. Two socks are randomly extracted from the basket. What is the probability of getting two socks of same color?

 Solution:

     The number of equally possible cases is the number of all 2-size combinations of socks, namely, C(6, 2) = 15. The number of favorable cases is three, because we have three pairs of socks having the same color. Thus, the probability is 3/15 = 1/5 = 20%.

3) An urn contains four white balls and six black balls. Two balls are drawn simultaneously. Find the probability of the events: a) A – drawing two white balls;  b) B – drawing two black balls;  c) C – drawing two balls of the same color.

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4) Two dice, one red and one blue, are rolled. Consider the events: A – occurrence of a number less than 4 on the red die;  B – occurrence of a number less than 3 on the blue die. Find P(A or B).

Solution:

     The cases that are favorable to A are {1}, {2} and {3}; therefore, P(A) = 3/6. The cases that are favorable to B are {1} and {2}; therefore,P(A) = 2/6.

     The cases that are favorable to A and B correspond to the ordered pairs (1, 1), (1, 2), (2, 1),

(2, 2), (3, 1), (3, 2), and total six, in a probability field where the number of equally possible cases is 6 x 6 = 36. We then have P(A and B) = 6/36. The requested probability is

[pic]

5) At a blackjack game, calculate the probability for a player to get a total of twenty points from the first two cards (provided no other cards are shown), if a 52-card deck is used.

 Solution:

     The variants totaling twenty points are of the type A + 9 or 10 + 10 (as a value; that is, any 2-size combination of cards from 10, J, Q, K). We have sixteen variants A + 9 (4 aces and 4 nines) and C(16, 2) = 120 variants 10 + 10 (all 2-size combinations of cards from the sixteen cards with a value of 10). The number of all possible distribution variants for two cards is C(52, 2)=1326 . The probability is then P = (16 + 120)/1326 = 68/663.

6) We have two urns, the first containing three white balls and four black balls and the second three white balls and five black balls. A ball is drawn from a randomly chosen urn. Find the probability for the drawn ball to be white.

 Solution:Denote the events: A – the first urn is the chosen one; B – the second urn is the chosen one; C – the drawn ball is white. A and B form a complete system of events and P(A) = P(B) = 1/2.

     We have  P(C│A) = 3/7 and  P(C│B) = 3/8.  According to total probability formula, we have:

     P(C) = P(A)P(C│A) +P(B)P(C│B) = (1/2) x (3/7) + (1/2) x (3/8) = 45/112 = 0.40178.

7) Five cards are drawn at once from a 32-card deck. What is the probability of the five cards containing at least one queen (Q)?

 Solution:

     Denoting by A the event to be measured the five extracted cards contain at least one Q, we then calculate the probability of the contrary event A [pic]– the five extracted cards contain no Q.

     The equally possible elementary events are the occurrences of 5-size combinations of cards from the 32, a total of C(32, 5). The combinations that are favorable to event A [pic] have the form (xyztv), with x, y, z, t, v taking any card as value, except the four Q-cards. They total C(32 - 4, 5) = C(28, 5).

     We then have:

    [pic]

 

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Count operation

Select part of object

Select all object

With order

Without order

Without order

With order

Without replication

With replication

1

N!/(n1!.n2!...nk!)

N!

At the same time

Sequentially

At the same time

Sequentially

Com (n taken r)

Per (n taken r)

Without replacement

With replacement

Without replacement

With replacement

com (n taken 1). com (n-1 taken 1)…..((r times))

Per (n taken 1). Per (n-1 taken 1)…..((r times))

Per (n taken 1). Per (n taken 1)…..((r times))

com (n taken 1). com (n-1 taken 1)…..((r times))

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