Math 3680 Lecture #5 Important Discrete Distributions

The Binomial Distribution

Example: A student randomly guesses at three questions. Each question has five possible answers, only once of which is correct. Find the probability that she gets 0, 1, 2 or 3 correct. This is the same problem as the previous one; we will now solve it by means of the binomial formula.

Example: Recall that if X ~ Binomial(3, 0.2), P(X = 0) = P(X = 1) = P(X = 2) = P(X = 3) = Compute E(X) and SD(X).

MOMENTS OF Binomial(n,  ) DISTRIBUTION E(X) = n  SD(X) = Var(X) = Try this for the Binomial(3, 0.2) distribution. Do these formulas make intuitive sense?

Example: A die is rolled 30 times. Let X denote the number of aces that appear. A) Find P(X = 3). B) Find E(X) and SD(X).

Example: Three draws are made with replacement from a box containing 6 tickets: two labeled “1”, one each labeled, “2”, “3”, “4” and “5”. Find the probability of getting two “1”s.

The Hypergeometric Distribution

Example: Three draws are made without replacement from a box containing 6 tickets: two labeled “1”, one each labeled, “2”, “3”, “4” and “5”. Find the probability of getting two “1”s.

P(S 1 S 2 S 3 ) = P(S 1 ) P(S 2 | S 1 ) P(S 3 | S 1 ∩ S 2 ) = P(S 1 S 2 F 3 ) = P(S 1 ) P(S 2 | S 1 ) P(F 3 | S 1 ∩ S 2 ) = P(S 1 F 2 S 3 ) = P(S 1 ) P(F 2 | S 1 ) P(S 3 | S 1 ∩ F 2 ) = P(S 1 F 2 F 3 ) = P(S 1 ) P(F 2 | S 1 ) P(F 3 | S 1 ∩ F 2 ) = P(F 1 S 2 S 3 ) = P(F 1 ) P(S 2 | F 1 ) P(S 3 | F 1 ∩ S 2 ) = P(F 1 S 2 F 3 ) = P(F 1 ) P(S 2 | F 1 ) P(F 3 | S 1 ∩ F 2 ) = P(F 1 F 2 S 3 ) = P(F 1 ) P(F 2 | F 1 ) P(S 3 | F 1 ∩ F 2 ) = P(F 1 F 2 F 3 ) = P(F 1 ) P(F 2 | F 1 ) P(F 3 | F 1 ∩ F 2 ) =

The Hypergeometric Distribution Suppose that n draws are made without replacement from a finite population of size N which contains G “good” objects and B = N - G “bad” objects. Let X denote the number of good objects drawn. Then where b = n - g.

Example: Three draws are made without replacement from a box containing 6 tickets; two of which are labeled “1”, and one each labeled, “2”, “3”, “4” and “5”. Find the probability of getting two “1”’s.

MOMENTS OF HYPERGEOMETRIC(N, G, n) DISTRIBUTION E(X) = n  where  = G / N) SD(X) = Var(X) =

REDUCTION FACTOR The term is called the Small Population Reduction Factor. It always appears when we draw without replacement. If the population is large (N > 20 n), then the reduction factor can generally be ignored (why?).

Example: Thirteen cards are dealt from a well-shuffled deck. Let X denote the number of hearts that appear. A) Find P(X = 3). B) Find E(X) and SD(X).

The Poisson Distribution

Example. A lonely bachelor decides to play the field, deciding that a lifetime of watching “Leave It To Beaver” reruns doesn’t sound all that pleasant. On 250 consecutive days, he calls a different woman for a date. Unfortunately, through the school of hard knocks, he knows that the probability that a given woman will accept his gracious invitation is only 1%. What is the chance that he will land three dates?

The Poisson Distribution If X is a discrete random variable that satisfies for  > 0, then we say that X has a Poisson(  ) distribution. Property: This is a reasonable approximation of the Binomial(n,  ) distribution if n > 100 and n  < 6. (In other words, n is large and  is very small.)

Exercise: Confirm that is a probability distribution for  > 0.

MOMENTS OF THE POISSON(  ) DISTRIBUTION E(X) =  SD(X) = Var(X) =  (In terms of the binomial distribution, why?)

Example: It is estimated that one in two thousand spectators at a sporting event will require first aid treatment. Suppose that there are 11,000 fans attending a particular event. Find the probability that at least three spectators will require treatment.

The Poisson Process

A process is an experiment where events occur at random times, like the counts of a Geiger counter detecting radioactive decay. Although the decays occur at random times, the process seems to satisfy three conditions. Loosely stated, they are: For a very short time interval, if we double the length of time, the probability of a decay during the time interval will double. In a short time interval, we are very unlikely to observe two or more decays. Just because we detected a decay during a one-second time interval, we do not expect it to be any more or less likely that we will detect another decay in the next one-second time interval. If a process satisfies these three conditions, then we call it a Poisson process.

Events like mutations in a population and earthquakes can be modeled as a Poisson process. However, a Poisson process may not be an appropriate model for other events. For instance, cars on a highway tend to cluster behind a slowly moving car. If an event is the passing of a car, then the third condition is not satisfied. This is because if a car just passed, then it is likely that more passes from the other cars in the cluster will occur in the near future.

The Poisson Process Suppose that a Poisson process has arrivals that occur at rate per second (or other unit). Then the number of arrivals that occur in any interval with length t seconds has a Poisson( t ) distribution.

Example. In the Luria-Delbrück mutation model, it is assumed that the occurrences of mutations follow a Poisson process with an average of 0.25 mutations per hour. Find the probability that at least one mutation occurs in the next two hours.