Binomial Table for n=7, n=8 and n=9

A binomial random variable provides an important example of a discrete random variable.  The binomial distribution, which describes the probability for each value of our random variable, can be determined completely by the two parameters: n  and p.  Here n is the number of independent trials and p is the constant probability of success in each trial.  The tables below provide binomial probabilities for n = 7,8 and 9.  The probabilities in each are rounded to three decimal places.

Should a  binomial distribution be used?.   Before jumping in to use this table, we need to check that the following conditions are met:

1. We have a finite number of observations or trials.
2. The outcome of each trial can be classified as either a success or a failure.
3. The probability of success remains constant.
4. The observations are independent of one another.

When these four conditions are met, the binomial distribution will give the probability of r successes in an experiment with a total of n independent trials, each having probability of success p.   The probabilities in the table are calculated by the formula C(n, r)p^r(1 - p)^{n - r} where C(n, r) is the formula for combinations.  There are separate tables for each value of n.  Each entry in the table is organized by the values of p and of r. 

Other Tables

For other binomial distribution tables we have n = 2 to 6, n = 10 to 11. When the values of np and n(1 - p) are both greater than or equal to 10, we can use the normal approximation to the binomial distribution.  This gives us a good approximation of our probabilities and does not require the calculation of binomial coefficients.  This provides a great advantage because these binomial calculations can be quite involved.

Example

Genetics has many connections to probability.  We will look at one to illustrate the use of the binomial distribution.  Suppose we know that probability of an offspring inheriting two copies of a recessive gene (and hence possessing the recessive trait we are studying) is 1/4. 

Furthermore, we want to calculate the probability that a certain number of children in an eight-member family possesses this trait.  Let X be the number of children with this trait.  We look at the table for n = 8 and the column with p = 0.25, and see the following:

.100
.267.311.208.087.023.004

This means for our example that

• P(X = 0) = 10.0%, which is the probability that none of the children has the recessive trait.
• P(X = 1) = 26.7%, which is the probability that one of the children has the recessive trait.
• P(X = 2) = 31.1%, which is the probability that two of the children have the recessive trait.
• P(X = 3) = 20.8%, which is the probability that three of the children have the recessive trait.
• P(X = 4) = 8.7%, which is the probability that four of the children have the recessive trait.
• P(X = 5) = 2.3%, which is the probability that five of the children have the recessive trait.
• P(X = 6) = 0.4%, which is the probability that six of the children have the recessive trait.

Tables for n = 7 to n = 9

n = 7

n = 8

n = 9