15'1 Probability and Simple Experiments

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15.1 Probability and Simple Experiments

• 1. Experiment- Act of making an observation or
  taking a measurement.
• 2. Outcome- One of possible things that can occur
  as a result of an experiment.
• 3. Sample Space- The set of all possible outcomes
  is called the sample space.
• 4. Event- An event is any subset of the sample
  space

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• Determine the sample space for the experiment
  below- Rolling a standard six sided die with
  1,2,3,4,5,6 dots respectively, on the six faces
  and recording the number of dots showing on the
  top
• 2. What is the event of getting an even number of
  dots on a roll

1. S 1,2,3,4,5,6 2. E2,4,6
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• Determine the sample space for the experiment
  below- Rolling two standard six sided die with
  1,2,3,4,5,6 dots respectively, on the six faces
  and recording the number of dots showing on the
  top
• 2. What is the event of getting a sum of even
  number of dots on a roll

1. S (1,1), (1,2), (1,3) (1,6), (2,1), (2,2),
(2,6)(6,1),(6,2),(6,6)
2. E(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),
(3,3), (3,5), (4,2), (4,4),(4,6), (5,1),(5,3),
(5,5),(6,2),(6,4),(6,6)
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Computing Probability in Simple Experiments

• Probability of an event with equally likely
  outcomes- Let E be an event, n(E)number of
  elements in E, and S be the sample space with
  n(S)number of elements in S, then the
  probability of event E occurring is
• P(E)

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• Example-Determine the probability of getting a
  sum of even number of dots on a roll.

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Find the probability of picking a red marble from
the box below-
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• Complement of the event E is written as
  and

2. Two events A and B are mutually exclusive if
events A and B are disjoint
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• 3. P(AUB)P(A)P(B) for mutually exclusive
  events.
• 4.P(AUB)P(A)P(B)-P(A?B)

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Example

• Consider rolling 2 standard dies. Find the
  probability of getting a sum of 2 and then the
  probability that the sum is 8. Are the two events
  mutually exclusive and if they are, find the
  probability of the union of the two events.
• n(S)62, Agetting a sum of 1(1,1,)
• n(A)1, P(A)1/62
• Bgetting a sum of 8

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• 82662, 83553, 844, these are all
  different ways to add to 8. This gives us 5 ways
  therefore
• n(B)5 and P(B)5/625/36
• P(AUB)

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• Ex 2. Find the probability of picking a red
  marble from a box containing 4 red marbles, 8
  green marbles and 2 white marbles.
• Soln- Epicking red marble, n(E)4, P(E)

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• A box contains 4 red marbles, 8 green marbles and
  2 white marbles. Consider the 3 events Apicking
  a red marble, Bpicking a green marble and
  Cpicking a white marble. Find P(A), P(B), P(C).
  What is P(A)P(B)P(C)?.

• Soln P(A) P(B) and
• P(C)

and so P(A)P(B)P(C)1
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• In the previous problem suppose that an unknown
  number of white marbles are added and it is known
  that P(A)

and P(B)
what is P(C)?

• Since P(A)P(B)P(C)1, we get P(C)1-P(A)-P(B)

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Probability and Complex Experiments

• Tree Diagram- Is used to represent the outcomes
  of an experiment.
• Ex. Consider the experiment of picking 2 marbles
  from a jar containing 2 white marbles, 3 green
  marbles and 1 red marble without replacement. We
  can represent this by the following tree

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Without Replacement
OUTCOMES
2nd draw
1st draw
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With Replacement
OUTCOMES
2nd draw
1st draw
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Find the probability of picking a white first and
then green (without replacement)
1st draw
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Find the probability of picking a white first and
then green (with replacement)
1st draw
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Property

• Additive Property of Probability Tree Diagram-
  Suppose that an event E is a union of pair wise
  mutually exclusive simpler events E1, E2,En,
  where E1, E2,En are events from a sample sample
  space, then P(E)P(E1)P(E2)P(En). The
  probabilities of the events E1, E2,En can be
  viewed as those associated with the ends of the
  branches in a probability tree diagram

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Probability with Permutations and Combinations

• Permutation- An ordered arrangement of
• objects is called a permutation.
• Theorem- The number of permutations
• of n distinct objects, taken all together is n!
• Example- Consider the three letters D, O,
• G, write all possible permutations- DOG, DGO,
  GOD, GDO, OGD, ODG
• Total 6 or 3!arrangement of each combination

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Choosing r objects out of n

• Theorem- The number of permutations of r objects
  chosen from n objects where 0?r ?n, is

• For example how many sequences with 3 letters
  (without repetition) can you have?
• Well there are 26 letter and so consider the
  three spots

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• or we have

Ex- How many 5 digit numbers can be formed using
1,2, 3,4,5,6,7,8,9 without any of the numbers
being repeated?


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• Ex. Certain automobile license plates consist of
  a sequence of three letters followed by three
  digits. If no repetitions of letters or numbers
  are allowed, how many license plates are possible?

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Combinations

• Combinations- A collection of objects, in no
  particular order, is called a combination.
• Example- Consider the four letters D, O,G,T write
  all possible combinations-
• D,O,G,D,O,T,D,G,T,O,G,T
• Total4 combinations or

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• that means that it is the total number of
  permutations divided by 3!, this occurs since in
  combinations order is not important and so when
  we count permutations where order is important,
  all the sets with three letters have 3! ways and
  that is the number of repeated combinations (for
  example D,O,G is the same as O,D,G).

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• Theorem- The number of combinations of r objects
  chosen from n objects, where 0?r ?n, is

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Pascals Triangle

• 1
• 1 1
• 1 2 1
• 1 3 3 1
• 1 4 6 4 1
• 1 5 10 10 5 1