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The Binomial Distribution

The binomial distribution is a discrete probability distribution based on a Bernoulli process.

Bernoulli Process:

Let X=the number of successes achieved in R trials of a Bernoulli process.  Then, we say that X is a binomial random variable and the distribution of X is the binomial distribution.

Example:  Consider the following binomial experiment.  Items from a particular lot are either found to be defective (D) or non-defective (N).  Suppose that the probability that an item is defective is 25%.  Also, suppose that whether an item is found to be defective is independent from item to item.  Select 3 items.  Then, there are 8 possible outcomes:


Let X be the number of defectives out of the 3 items chosen.  Then, X will have the following values:


We need to find the probability that X takes on each of those values.  Using the basic principles of probability, let's find the probability of exactly 1 defective (X=1).

Pr(X=1) = Pr(DNN) + Pr(NDN) + Pr(NND)
             = Pr(D)Pr(N)Pr(N) + Pr(N)Pr(D)Pr(N) + Pr(N)Pr(N)Pr(D)
             = 3*(0.25)1(1-0.25)3-1 = 27/64

If we let

n=# of trials
r=# of successes out of n trials
pi=probability of success

we could rewrite this expression as  Pr(X=1) = 3*pir(1-pi)n-r
and the (3) in this expression is just the number of possible ways to choose r objects out of n.  So, if we had a formula for determining how many ways you can choose r objects out of n, then we could write this expression in a completely generic manner.

Cnr = # of possible combinations of r objects from n objects = n!/{r!(n-r)!}

So to calculate the probability of achieving r successes out of n trials:

b(r;n,pi) = Pr(X=r) = Cnr pir(1-pi)n-r  for r=0,1,2,...,n


The probability that a certain kind of component will survive a given shock test is 3/4.  Find the probability that exactly 2 of the next 4 components survive.


For this example, survival is a success.  The probability of a success is pi=3/4.  The number of trials is n=4, and the number of successes is r=2.  So, if X=# of items that survive out of the next 4, then we find our probability as follows:

Pr(X=2) = b(2; 4, 3/4) = 4!/{2!2!} (3/4)2 (1 - 3/4)4-2 = (6)(9/16)(1/16) = 27/128


A traffic control engineer reports that 75% of the vehicles passing through a checkpoint are from within the state.  What is the probability that 3 of the next 9 vehicles are from within the state?  What is the probability that fewer than 4 of the next nine vehicles are from within the state?


Let X=# of vehicles out of the next 9 that are from within the state.  Then,  pi=prob. that a vehicle is from within the state = 0.75 and n=9.

What is the probability that 3 out of the next 9 are from within state?

Pr(X=3) = b(3; 9, 0.75) = 9!/{3!6!} (0.75)3 (1-0.75)9-3

What is the probability that fewer than 4 of the next 9 are from within state?

Pr(X<4) = Pr(X=0) + Pr(X=1) + Pr(X=2) + Pr(X=3) = sum{x=0 to 3} b(x; 9, 0.75)

        = 9!/{0!9!} (0.75)0 (1-0.75)9-0 + 9!/{1!8!} (0.75)1 (1-0.75)9-1  + 9!/{2!7!} (0.75)2 (1-0.75)9-2 + 9!/{3!6!} (0.75)3 (1-0.75)9-3

Calculating the probability in the previous example can be simplified by using the Cumulative Probability Distribution which has been tabulated in Table A.

Cumulative Probability Distribution:

B(r; n, pi) = Pr(X<=r) = sum{x=0 to r} b(x: n, pi)


The probability that a patient recovers from a rare blood disease is 0.4.  If 20 people are known to have contracted the disease, what is the probability that (a) exactly 5 of them survive, (b) from 3 to 8 of them survive, and (c) at least 10 of them survive?


(a)  Pr(X=5) = 20!/{15!5!} (0.4)5(1-0.4)15 = 0.0746

Alternatively, you can use Table A as follows to calculate the same probability:

  Pr(X=5) = Pr(X<=5) - Pr(X<=4) = (from Table A) 0.1256 - 0.0510 = 0.0746

(b) Pr(3<=X<=8) = Pr(X<=8) - Pr(X<3) = Pr(X<=8) - Pr(X<=2) = (from Table A) 0.5956 - 0.0036 = 0.5920

(c) Pr(X>=10) = 1 - Pr(X<10) = 1 - Pr(X<=9) = (from Table A) 1 - 0.7553 = 0.2447