1. DISCRETE PROBABILITY MODELS
BINOMIAL PROBABILITY MODEL
GEOMETRIC PROBABILITY MODEL
HYPERGEOMETRIC PROBABILITY MODEL

2. 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.

3. 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

4. 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

5. TOSSING A FAIR COIN 20 TIMES
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?

6. 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

7. 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.

8. 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.

9. 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.
A fair coin is tossed ten times. Let X be the number of times the coin lands heads
Five basketball players each attempt a free throw. Let Y be the number of free throws made.
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.

10. 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.

11. BINOMIAL PROBABILITY MODEL
LET n = NUMBER OF TRIALS
p = PROBABILITY OF SUCCESS
q = PROBABILITY OF FAILURE
X = NUMBER OF SUCCESSESS IN n TRIALS

12. n! = n(n-1)(n-2)(n-3) … 3.2.1

13. EXAMPLES
COMPUTE
(1) 3! (2) 4! (3) 5! (4) 6!

14. 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

15. EXAMPLE
AN OLYMPIC ARCHER IS ABLE TO HIT THE BULL’S-EYE 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? BULL’SEYES
(6) WITH WHAT STANDARD DEVIATION? 0.98

16. 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.

17. CONDITIONS SATISFYING A GEOMETRIC PROBABILITY MODEL
SAME AS THOSE FOR A BINOMIAL PROBABILITY MODEL

18. 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

19. 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

20. EXAMPLE
AN OLYMPIC ARCHER IS ABLE TO HIT THE BULL’S-EYE 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 =
IF SHE KEEPS SHOOTING ARROWS UNTIL SHE HITS THE BULL’S-EYE, HOW LONG DO YOU EXPECT IT WILL TAKE? ANS = 1.25 SHOTS

21. 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:
A RANDOM SAMPLE OF SIZE r IS SELECTED WITHOUT REPLACEMENT FROM n ITEMS.
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.

22. 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

23. PROBABILITY FUNCTION , f(x), OF A HYPERGEOMETRIC PROBABILITY MODEL

24. MEAN (EXPECTED VALUE) OF A RANDOM VARIABLE X DISTRIBUTED HYPERGEOMETRICALLY  

25. VARIANCE OF A RANDOM VARIABLE X DISTRIBUTED HYPERGEOMETRICALLY

26. STANDARD DEVIATION OF A RANDOM VARIABLE X DISTRIBUTED HYPERGEOMETRICALLY

27. 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.

28. 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

29. EXTENSION OF THE HYPERGEOMETRIC PROBABILITY MODEL

30. 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.