# Special Discrete Distributions

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Special Discrete Distributions
Chapter 7 Special Discrete Distributions

Binomial Distribution B(n,p)
Each trial results in one of two mutually exclusive outcomes. (success/failure) There are a fixed number of trials Outcomes of different trials are independent The probability that a trial results in success is the same for all trials The binomial random variable x is defined as the number of successes out of the fixed number

Are these binomial distributions?
Toss a coin 10 times and count the number of heads Yes Deal 10 cards from a shuffled deck and count the number of red cards No, probability does not remain constant Two parents with genes for O and A blood types and count the number of children with blood type O No, no fixed number

Binomial example Take the example of 5 coin tosses. What’s the probability that you flip exactly 3 heads in 5 coin tosses?

Binomial distribution
Solution: One way to get exactly 3 heads: HHHTT What’s the probability of this exact arrangement? P(heads)xP(heads) xP(heads)xP(tails)xP(tails) =(1/2)3 x (1/2)2 Another way to get exactly 3 heads: THHHT Probability of this exact outcome = (1/2)1 x (1/2)3 x (1/2)1 = (1/2)3 x (1/2)2

Binomial distribution
In fact, (1/2)3 x (1/2)2 is the probability of each unique outcome that has exactly 3 heads and 2 tails. So, the overall probability of 3 heads and 2 tails is: (1/2)3 x (1/2)2 + (1/2)3 x (1/2)2 + (1/2)3 x (1/2)2 + ….. for as many unique arrangements as there are—but how many are there??

5C3 = 5!/3!2! = 10 Outcome Probability THHHT (1/2)3 x (1/2)2
HHHTT (1/2)3 x (1/2)2 TTHHH (1/2)3 x (1/2)2 HTTHH (1/2)3 x (1/2)2 HHTTH (1/2)3 x (1/2)2 THTHH (1/2)3 x (1/2)2 HTHTH (1/2)3 x (1/2)2 HHTHT (1/2)3 x (1/2)2 THHTH (1/2)3 x (1/2)2 HTHHT (1/2)3 x (1/2)2 10 arrangements x (1/2)3 x (1/2)2 The probability of each unique outcome (note: they are all equal) ways to arrange 3 heads in 5 trials 5C3 = 5!/3!2! = 10

Binomial distribution function: X= the number of heads tossed in 5 coin tosses
p(x) p(x) x 1 2 3 4 5 number of heads number of heads

Pascal’s Triangle!

Example 2 As voters exit the polls, you ask a representative random sample of 6 voters if they voted for proposition 100. If the true percentage of voters who vote for the proposition is 55.1%, what is the probability that, in your sample, exactly 2 voted for the proposition and 4 did not?

Solution: . 15 arrangements x (.551)2 x (.449)4
Outcome Probability YYNNNN = (.551)2 x (.449)4 NYYNNN (.449)1 x (.551)2 x (.449)3 = (.551)2 x (.449)4 NNYYNN (.449)2 x (.551)2 x (.449)2 = (.551)2 x (.449)4 NNNYYN (.449)3 x (.551)2 x (.449)1 = (.551)2 x (.449)4 NNNNYY (.449)4 x (.551) = (.551)2 x (.449)4 . ways to arrange 2 Obama votes among 6 voters 15 arrangements x (.551)2 x (.449)4   P(2 yes votes exactly) = x (.551)2 x (.449)4 = 18.5%

Binomial Formula: n = number of trials 1-p = probability of failure
K = # successes out of n trials Where: p = probability of success

Toss a 3 coins and count the number of heads
Find the discrete probability distribution X P(x) Out of 3 coins that are tossed, what is the probability of getting exactly 2 heads?

Out of 3 coins that are tossed, what is the probability of getting exactly 2 heads?

Binomial distribution: example
If I toss a coin 20 times, what’s the probability of getting exactly 10 heads?

The number of inaccurate gauges in a group of four is a binomial random variable. If the probability of a defect is 0.1, what is the probability that only 1 is defective? More than 1 is defective?

Practice problems If the probability of being a smoker among a group of cases with lung cancer is .6, what’s the probability that in a group of 8 cases you have less than 2 smokers? More than 5? What are the expected value and variance of the number of smokers?

Answer 1 1 1 1 2 1

Answer, continued 1 4 5 2 3 6 7 8

Answer, continued 1 4 5 2 3 6 7 8 E(X) = 8 (.6) = 4.8
P(>5)= = .3168 P(<2)= = E(X) = 8 (.6) = 4.8 Var(X) = 8 (.6) (.4) =1.92 StdDev(X) = 1.38

Calculator Binomialpdf(n,p,x) – this calculates the probability of a single binomial P(x = k) Binomialcdf(n,p,x) – this calculates the cumulative probabilities from P(0) to P(k) OR P(X < k)

A genetic trait of one family manifests itself in 25% of the offspring
A genetic trait of one family manifests itself in 25% of the offspring. If eight offspring are randomly selected, find the probability that the trait will appear in exactly three of them. Less than half? At least 5?

P(x < 6) = binomcdf(10,.3,6) = .9894
In a certain county, 30% of the voters are Republicans. If ten voters are selected at random, find the probability that no more than six of them will be Republicans. P(x < 6) = binomcdf(10,.3,6) = .9894 What is the probability that at least 7 are not Republicans? P(x > 7) = 1 - binomcdf(10,.7,6) = .6496

Binomial formulas for mean and standard deviation

In a certain county, 30% of the voters are Republicans
In a certain county, 30% of the voters are Republicans. How many Republicans would you expect in ten randomly selected voters? What is the standard deviation for this distribution? expect

What happens to the shape of the distribution as the probability of success increased?

Sketch histogram on board
Binomial Activity In L1 – seq(x,x,0,10) In L2 – binompdf(10, .1 ,L1) Sketch histogram on board

Binomial Activity What happened to the shape of the distribution as the probability of success increased? As the probability of success increases, the shape changes from being skewed right to symmetrical at p =.5 to skewed left.

Binomial Activity Calculate the mean and standard deviations for each of the probabilities What do you notice? As the probability of success increases, the means increase. the standard deviations increase to p = .5, then decrease. Their values are also symmetrical.

Geometric Distributions:
There are two mutually exclusive outcomes Each trial is independent of the others The probability of success remains constant for each trial. The random variable x is the number of trials UNTIL the FIRST success occurs. So what are the possible values of X How far will this go? To infinity X 1 2 3 4 . . .

Differences between binomial & geometric distributions
The difference between binomial and geometric properties is that there is NOT a fixed number of trials in geometric distributions!

Other differences: Binomial random variables start with 0 while geometric random variables start with 1 Binomial distributions are finite, while geometric distributions are infinite

Not on formula sheet – they will be given on quiz or test
Geometric Formulas: Not on formula sheet – they will be given on quiz or test

What are the values for these random variables?
Count the number of boys in a family of four children. What are the values for these random variables? Binomial: X Count children until first son is born Geometric: X

No “n” because there is no fixed number!
Calculator geometpdf(p,x) – finds the geometric probability for P(X = k) Geometcdf(p,x) – finds the cumulative probability for P(X < k) No “n” because there is no fixed number!

What is the probability that the first son is the fourth child born?
What is the probability that the first son born is at most the fourth child? What is the probability that the first son born is at least the third child?

A real estate agent shows a house to prospective buyers
A real estate agent shows a house to prospective buyers. The probability that the house will be sold to the person is 35%. What is the probability that the agent will sell the house to the third person she shows it to? How many prospective buyers does she expect to show the house to before someone buys the house? SD?

In L1 – input the x-values In L2 – geometpdf(.1 ,L1)
Geometric Activity In L1 – input the x-values In L2 – geometpdf(.1 ,L1) Sketch histogram on board

Geometric Activity What happened to the shape of the distribution as the probability of success increased? As the probability of success increases, the shape becomes more & more strongly skewed right

Geometric Activity What do you notice?
Calculate the mean and standard deviations for each of the probabilities What do you notice? As the probability of success increase, the means decrease the standard deviations decrease.

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