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Distributions Counting the Number of Discrete Events

The Binomial distribution

Gives the probability of exactly k successes in n trials, where p = probability of a success on any given trail.


For example, the probability of no successes is

• Prob( k = 0) = (1-p)ⁿ

Likewise, the probability of all successes is

• Prob( k = n) = pⁿ


Derivation

What is the probability of one success in five traits where Prob(success) = p? Let 1 = success, 0 = failure.



Thus, the overall probability is 5 * p (1-p)⁴.

More generally, consider the probability of obtaining k successes in n total trials.

• The probability that the first k trials are all successes is p^k and the probability that the last (n-k) trials are all failures is (1-p)^n-k, for a total probability of p^k(1-p)^n-k.

• There are a total of n!/[ k!(n-k)! ] such combinations of getting k success in n trials, each of which has probability p^k(1-p)^n-k.

Example 1

• Assuming Pr(girl) = 1/2, Pr(all 7 are girls) = (1/2)⁷ = 0.0078

The Poisson distribution

Letting = expected number of successes, then the probability that we observe k successes is

In particular,

The Poisson and binomial are connected in that = np, so that if n is large (i.e., n >> ), the Poisson provides a quick approximation of the binomial.

The Poisson distribution follows by assuming that the expected number of events occurring in some small interval of time is

so that the rate of events per time unit is

Hence, the Poisson assumes that there is a constant rate of successes occurring, which is akin to assuming independent trials.

Example 2

Example 3

Summary: Binomial vs. Poisson

Distribution	Known parameters
Binomial	Number of Trials, n Probability of success/trail, p
Poisson	Expected number of successes, np =

Note that if n >> , we can set = np and use the Poisson to approximate the Binomial.

Waiting Time Distributions

How long until the first success?

The Geometric distribution

Let p = Probability of a success. If all trials are independent, then

Hence, while the average number of trials required is 1/p, there is a distribution about this average value.

Example 4

Suppose both parents are Aa, where the aa genotype displays a horrific disease, and as a consequence the family stops having children after the first such child appears. What is the probability they will have 1, 2, 3, or 4 children? Since Pr( aa)= 1/4 ,

Here is the distribution for up to 12 children

The Exponential distribution

Gives the waiting time to the first success when time is now continuous (i.e., how long until your light bulb fails?) Here the distribution parameter is , the success rate per time interval. Hence, the mean time until a success is 1/.

The exponential distribution is very closely related to the geometric, with and p having essentially the same role. Noting that

the geometric probability can be approximated by

Under the exponential distribution, the waiting time probabilities are given by the appropriate area under the curve given by

e^{- t}

Thus, the probability that the first success occurs at, or before, time T is

Example 5

Suppose you have a constant risk of 0.01 per year of getting cancer. This gives an average age for getting cancer at age 100. What is the probability you are cancer-free at ages 20, 40, 60, and 80?

giving

Here is the distribution out to age 200

The Normal Distribution

The parameters for this distribution are the mean and the variance (a measure of the spread). The square root of the variance, , is referred to as the standard deviation.

As the figure shows, the mean corresponds to the peak of the distribution, while the variance measures the spread. The larger the variance, the more spread out the distribution is.

The probability of a particular event is just given by the area under the normal (bell-shaped) curve.

Under the normal distribution, 95 percent of all values lie within 1.95 Standard Deviations of the mean,

Normal approximation of the binomial

For n large and p moderate, number of successes roughly follows a normal distribution with mean = np and variance = np(1-p).

Thus, approximately 95% of the values fall within the interval

Example

The probability that in individual is a certain genotype is expected to be 0.05. Should we be suspicious if we observe 70 such genotypes in a population of 1000 individuals? Using the normal approximation with n = 1000 and p = 0.05, = np = 50, = np(1-p) = 47.5. Since the upper 95% limit is + 1.96* = 63.5, on average we expect such an excess of genotypes to occur less than 5% of the time. Hence, we should indeed be suspicious.

Onto: Lecture 3