Probability Distribution. Binomial Distribution. Probability of Binary Events. Probability of success = p p(success) = p Probability of failure = q p(failure) = q p+q = 1 q = 1-p. Binomial Probability Distribution. A fixed number of observations (trials), n
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Take the example of 5 coin tosses. What’s the probability that you flip exactly 3 heads in 5 coin tosses?
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)3x(1/2)2
Another way to get exactly 3 heads: THHHT
Probability of this exact outcome = (1/2)1x (1/2)3x(1/2)1 = (1/2)3x(1/2)2
In fact, (1/2)3x(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)3x(1/2)2 + (1/2)3x(1/2)2+ (1/2)3x(1/2)2+ ….. for as many unique arrangements as there are—but how many are there??
10 arrangements x (1/2)3x(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
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?
ways to arrange 2 Obama votes among 6 voters
P(2 yes votes exactly) = x (.551)2x(.449)4= 18.5%
YYNNNN = (.551)2x(.449)4
NYYNNN(.449)1x (.551)2x(.449)3= (.551)2x(.449)4
NNYYNN(.449)2x (.551)2x(.449)2= (.551)2x(.449)4
NNNYYN(.449)3x (.551)2x(.449)1= (.551)2x(.449)4
NNNNYY(.449)4x (.551)2= (.551)2x(.449)4
15 arrangements x (.551)2x(.449)4
n = number of trials
1-p = probability of failure
p = probability of success
X = # successes out of n trials
Note the general pattern emerging if you have only two possible outcomes (call them 1/0 or yes/no or success/failure) in n independent trials, then the probability of exactly X “successes”=
Bernouilli trial: If there is only 1 trial with probability of success p and probability of failure 1-p, this is called a Bernouilli distribution. (special case of the binomial with n=1)
Probability of success:
Probability of failure:
For a Poisson random variable, the variance and mean are the same!
where = expected number of hits in a given time period
The Poisson distribution is used to model the number of events occurring within a given time interval. The formula for the Poisson probability density (mass) function is
is the shape parameter which indicates the average number of events in the given time interval.
Arrivals at a bus-stop follow a Poisson distribution with an average of 4.5 every quarter of an hour.
Obtain a barplot of the distribution (assume a maximum of 20 arrivals in a quarter of an hour) and calculate the probability of fewer than 3 arrivals in a quarter of an hour.
The probabilities of 0 up to 2 arrivals can be calculated directly from the formula
So p(0) = 0.01111
Similarly p(1)=0.04999 and p(2)=0.11248
So the probability of fewer than 3 arrivals is 0.01111+ 0.04999 + 0.11248 =0.17358
The Poisson distribution models counts, such as the number of new cases of SARS that occur in women in New England next month.
The distribution tells you the probability of all possible numbers of new cases, from 0 to infinity.
If X= # of new cases next month and X ~ Poisson (), then the probability that X=k (a particular count) is:
The mean and variance are both equal to .
The sum of independent Poisson variables is a further Poisson variable with mean equal to the sum of the individual means.
As well as cropping up in the situations already mentioned, the Poisson distribution provides an approximation for the Binomial distribution.
Simulation“Discrete-Event System Simulation”
Statistical Models in Simulation
• Some statistical model might well describe the variations.
• Select a known distribution through educated guesses
• Make estimate of the parameters
• Test for goodness of fit
• Discrete random variables
• Cumulative distribution function
• Queueing systems
• Inventory and supply-chain systems
• Reliability and maintainability
• Limited data
• Exponential distribution: if service times are completely random
• Normal distribution: fairly constant but with some random variability
(either positive or negative)
• Truncated normal distribution: similar to normal distribution but with restricted value.
• Gamma and Weibull distribution: more general than exponential
• The number of units demanded per order or per time period
• The time between demands
• The lead time = Time between placing an order and the receipt of that order
• Poisson: simple and extensively tabulated.
• Negative binomial distribution: longer tail than Poisson (more large demands).
• Geometric: special case of negative binomial given at least one demand has occurred.
• Exponential: failures are random
• Gamma: for standby redundancy where each component has an exponential TTF
• Weibull: failure is due to the most serious of a large number of defects in a system of components
• Normal: failures are due to wear