Transcript OF A HYPERGEOMETRIC PROBABILITY MODEL

DISCRETE PROBABILITY MODELS
• BINOMIAL PROBABILITY MODEL
• GEOMETRIC PROBABILITY MODEL
• HYPERGEOMETRIC PROBABILITY
MODEL
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BINOMIAL PROBABILITY MODELS
OBJECTIVES
• DETERMINE WHETHER A RANDOM
VARIABLE OR RANDOM EXPERIMENT IS
BINOMIAL.
• DETERMINE THE PROBABILITY
DISTRIBUTION OF A BINOMIAL RANDOM
VARIABLE.
• COMPUTE BINOMIAL PROBABILITIES.
• COMPUTE THE MEAN, VARIANCE, AND
STANDARD DEVIATION OF A BINOMIAL
RANDOM VARIABLE.
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QUESTION
• WHAT DOES THE PREFIX BI MEAN?
• THE PREFIX BI MEANS “TWO.”
• WE WILL BE CONSIDERING RANDOM
EXPERIMENTS IN WHICH THERE ARE
ONLY TWO OUTCOMES.
• THE TWO OUTCOMES WILL BE
CALLED SUCCESS OR FAILURE
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BINOMIAL PROBABILITY MODELS
BERNOULLI TRIAL
A RANDOM EXPERIMENT WITH TWO
COMPLEMENTARY OUTCOMES, ONE CALLED
SUCCESS (S), AND THE OTHER CALLED
FAILURE (F), IS CALLED A BERNOULLI TRIAL.
P(SUCCESS) = p
P(FAILURE) = q = 1 - p
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EXAMPLES
• TOSSING A FAIR COIN 20 TIMES
SUCCESS = HEADS WITH p = 0.5 AND
FAILURE = TAILS WITH q = 1 – p = 0.5
NOTE
• EACH TOSS OF THE COIN IS A TRIAL
• LET THE RANDOM VARIABLE X BE THE
NUMBER OF SUCCESSES IN THE 20
TOSSES. IS X BERNOULLI?
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• PRODUCTS COMING OUT OF A PRODUCTION LINE
SUCCESS = DEFECTIVE ITEMS
FAILURE = NON-DEFECTIVE ITEMS
DEFINE Y THE NUMBER OF DEFECTIVE ITEMS
COMING OUT OF THE PRODUCTION LINE. IS Y
BERNOULLI?
• TAKING A MULTIPLE CHOICE EXAM UNPREPARED
SUCCESS = CORRECT ANSWER
FAILLURE = WRONG ANSWER
DEFINE Z THE NUMBER OF CORRECT ANSWERS
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CONDITIONS FOR A BINOMILA PROBABILITY
EXPERIMENT
(1) THE TRIALS MUST BE BERNOULLI, THAT IS, THE
RANDOM EXPERIMENT MUST HAVE TWO
COMPLEMENTARY OUTCOMES – SUCCESS OR
FAILURE;
(2) THE TRIALS MUST BE INDEPENDENT. THIS
MEANSTHE OUTCOME OF ONE TRIAL WILL NOT
AFFECT THE OUTCOME OF THE OTHER TRIAL.
(3) THE PROBABILITY OF SUCCESS IS THE SAME
FOR EACH TRIAL.
(4) THE NUMBER OF TRIALS IS FIXED OR THE
EXPERIMENT IF CONDUCTED A FIXED NUMBER OF
TIMES.
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REMARK ON INDEPENDENCE OF BERNOULLI
TRIALS
• THE 10% CONDITION
• BERNOULLI TRIALS MUST BE
INDEPENDENT. IF THAT ASSUMPTION IS
VIOLATED, IT IS STILL OKAY TO PROCEED
AS LONG AS THE SAMPLE IS SMALLER
THAN 10% OF THE POPULATION.
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EXAMPLES: DETERMINING WHETHER A RANDOM
VARIABLE IS BINOMIAL
• Determine which of the following are binomial
random variables. For those that are binomial, state
the two possible outcomes and specify which is a
success. Also state the values of n and p.
1. A fair coin is tossed ten times. Let X be the number
of times the coin lands heads
2. Five basketball players each attempt a free throw.
Let Y be the number of free throws made.
3. A simple random sample of 50 voters is drawn from
the residents in a large city. Let Z be the number who
plan to vote for a proposition to increase spending
on public schools.
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ARE YOU INTERESTED IN THE NUMBER OF
SUCCESSES IN BERNOULLI TRIALS?
• QUESTION: WHAT IS THE NUMBER OF
SUCCESSES IN A SPECIFIED NUMBER OF
TRIALS?
• THE BINOMIAL PROBABILITY MODEL
ANSWERS THIS QUESTION, THAT IS, THE
PROBABILITY OF EXACTLY k SUCCESSES IN
n TRIALS.
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BINOMIAL PROBABILITY MODEL
• LET n = NUMBER OF TRIALS
p = PROBABILITY OF SUCCESS
q = PROBABILITY OF FAILURE
X = NUMBER OF SUCCESSESS IN n TRIALS
 n  k nk
P( X  k )    p q
k 
n
n!
  
 k  k!(n  k )!
where,
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n! = n(n-1)(n-2)(n-3) … 3.2.1
E ( X )    np
SD( X )    npq
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EXAMPLES
• COMPUTE
(1) 3!
(2) 4!
• COMPUTE
5
(1)  
 2
(3) 5!
10 
(2)  
7 
(4) 6!
12 
(3)  
0 
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EXAMPLE
• ASSUME THAT 13% OF PEOPLE ARE LEFT-HANDED.
IF WE SELECT 5 PEOPLE AT RANDOM, FIND THE
PROBABILITY OF EACH OUTCOME BELOW.
• (1) THERE ARE EXACTLY 3 LEFTIES IN THE GROUP.
• 0.0166
• (2) THERE ARE AT LEAST 3 LEFTIES IN THE GROUP.
• 0.0179
• (3) THERE ARE NO MORE THAN 3 LEFTIES IN THE
GROUP. 0.9987
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EXAMPLE
• AN OLYMPIC ARCHER IS ABLE TO HIT THE BULL’SEYE 80% OF THE TIME. ASSUME EACH SHOT IS
INDEPENDENT OF THE OTHERS. IF SHE SHOOTS 6
ARROWS, WHAT’S THE PROBABILITY THAT
• (1) SHE GETS EXACTLY 4 BULL’S-EYES? 0.246
• (2) SHE GETS AT LEAST 4 BULL’S-EYES? 0.901
• (3) SHE GETS AT MOST 4 BULL’S-EYES? 0.345
• (4) SHE MISSES THE BULL’S-EYE AT LEAST ONCE?
•
0.738
• (5) HOW MANY BULL’S-EYES DO YOU EXPECT HER
TO GET?
4.8 BULL’SEYES
• (6) WITH WHAT STANDARD DEVIATION? 0.98
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GEOMETRIC PROBABILITY MODEL
• QUESTION: HOW LONG WILL IT TAKE TO
ACHIEVE THE FIRST SUCCESS IN A SERIES
OF BERNOULLI TRIALS?
• THE MODEL THAT TELLS US THIS
PROBABILITY (THAT IS, THE PROBABILITY
UNTIL FIRST SUCCESS) IS CALLED THE
GEOMETRIC PROBABILITY MODEL.
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CONDITIONS SATISFYING A GEOMETRIC
PROBABILITY MODEL
•SAME AS THOSE
FOR A BINOMIAL
PROBABILITY
MODEL
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GEOMETRIC PROBABILITY MODEL FOR
BERNOULLI TRIALS
• LET p = PROBABILAITY OF SUCCESS
AND q = 1 – p = PROBABILITY OF
FAILURE
X = NUMBER OF TRIALS UNTIL FIRST
SUCCESS OCCURS
x 1
P( X  x)  q
1
E( X )   
p
SD( X ) 
p
q
2
p
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EXAMPLE
• ASSUME THAT 13% OF PEOPLE ARE LEFT-HANDED. IF WE
SELECT 5 PEOPLE AT RANDOM, FIND THE PROBABILITY OF
EACH OUTCOME DESCRIBED BELOW.
• (1) THE FIRST LEFTY IS THE FIFTH PERSON CHOSEN?
0.0745
• (2) THE FIRST LEFTY IS THE SECOND OR THIRD PERSON.
0.211
• (3) IF WE KEEP PICKING PEOPLE UNTIL WE FIND A LEFTY, HOW
LONG WILL YOU EXPECT IT WILL TAKE?
7.69 PEOPLE
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EXAMPLE
• AN OLYMPIC ARCHER IS ABLE TO HIT THE BULL’SEYE 80% OF THE TIME. ASSUME EACH SHOT IS
INDEPENDENT OF THE OTHERS. IF SHE SHOOTS 6
ARROWS, WHAT’S THE PROBABILITY THAT
• (1) HER FIRST BULL’S-EYE COMES ON THE THIRD
ARROW? ANS = 0.032
• (2) HER FIRST BULL’S-EYE COMES ON THE
FOURTH OR FIFTH ARROW? ANS = 0.00768
• IF SHE KEEPS SHOOTING ARROWS UNTIL SHE
HITS THE BULL’S-EYE, HOW LONG DO YOU
EXPECT IT WILL TAKE? ANS = 1.25 SHOTS
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HYPERGEOMETRIC PROBABILITY
MODEL (DISTRIBUTION)
• WE WISH TO LOOK AT A PROBABILITY MODEL THAT DOES
NOT REQUIRE INDEPENDENCE AND IS BASED ON THE
SAMPLING DONE WITHOUT REPLACEMENT.
• A HYPERGEOMETRIC EXPERIMENT IS ONE THAT POSSESSES
THE FOLLOWING TWO PROPERTIES:
1. A RANDOM SAMPLE OF SIZE r IS SELECTED WITHOUT
REPLACEMENT FROM n ITEMS.
2. k OF THE n ITEMS MAY BE CLASSIFIED AS SUCCESSES AND
n – k ARE CLASSIFIED AS FAILURES.
THE NUMBER X OF SUCCESSES IN A HYPERGEOMETRIC
EXPERIMENT IS CALLED A HYPERGEOMETRIC RANDOM
VARIABLE.
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HYPERGEOMETRIC PROBABILITY MODEL
• Suppose we have a collection of n objects
that consists of n1 objects of one kind and n2
objects of another kind, so that n = n1 + n2. If
we take a random sample of r objects from
this collection without replacement and let X
equal the number of objects of the first kind
in the random sample, we say that X has a
hyper-geometric distribution. In this case,
the probability function of X is defined by
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PROBABILITY FUNCTION , f(x), OF A
HYPERGEOMETRIC PROBABILITY MODEL
•
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MEAN (EXPECTED VALUE) OF A RANDOM
VARIABLE X DISTRIBUTED
HYPERGEOMETRICALLY
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VARIANCE OF A RANDOM VARIABLE X
DISTRIBUTED HYPERGEOMETRICALLY
•
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STANDARD DEVIATION OF A RANDOM
VARIABLE X DISTRIBUTED
HYPERGEOMETRICALLY
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EXAMPLES
• 1. IF 13 CARDS ARE SELECTED AT RANDOM AND WITHOUT
REPLACEMENT OUT OF 52 CARDS, FIND THE PROBABILITY
THAT 6 ARE RED AND 7 ARE BLACK.
• 2. IF 5 CARDS ARE SELECTED AT RANDOM AND WITHOUT
REPLACEMENT OUT OF 52 CARDS, FIND THE PROBABILITY
OF SELECTING EXACTLY 2 HEARTS.
• 3. FROM AN URN THAT CONTAINS ONLY 7 ORANGE BALLS
AND 3 BLUE BALLS, TAKE A RANDOM SAMPLE OF 2 BALLS
WITHOUT REPLACEMENT. LET THE RANDOM VARIABLE X BE
THE NUMBER OF ORANGE BALLS IN THE SAMPLE. EFINE
THE PROBABILITY FUNCTION OF X.
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EXTENSION OF THE HYPERGEOMETRIC
DISTRIBUTION
• We can extend the hyper-geometric distribution to
determine probabilities for more than just two kinds
of objects in a collection.
• Suppose that a collection of n objects consists of k
different kinds of objects, n1 of one kind, n2 of
another kind, and on up to nk of a k – th kind. If we
take a random sample of r objects from this
collection and let X1 equal the number of objects of
the first kind, X2 equal the number of objects of the
second kind, and so on up to Xk objects of the k – th
kind, the joint probability function of these k random
variables is
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EXTENSION OF THE HYPERGEOMETRIC
PROBABILITY MODEL
•
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EXAMPLE
• If 13 cards are selected at random
and without replacement out of 52
cards, find the probability of
selecting 2 clubs, 5 diamonds, 4
hearts, and 2 spades.
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