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

Bernoulli Process:

- Series of trials, each of which results in a (0,1) outcome
- The probability of getting a 0 is the same for all trials.
- The outcomes of successive trials are independent.

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:

NNN

NDN

NND

NDD

DNN

DND

DDN

DDD

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

__ x__

0

1

2

3

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**pi*^{r}(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.

C^{n}_{r} = # 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) = C^{n}_{r} *pi*^{r}(1-*pi*)^{n-r}
for r=0,1,2,...,n

*Example*:

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.

*Solution*:

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

*Example:*

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?

*Solution:*

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*)

*Example*:

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?

*Solution:*

(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