Binomial Table for n = 2, 3, 4, 5 and 6

One important discrete random variable is a binomial random variable. The distribution of this type of variable, referred to as the binomial distribution, is completely determined by two parameters: and p.  Here n is the number of trials and p is the probability of success. The tables below are for n = 2, 3, 4, 5 and 6. The probabilities in each are rounded to three decimal places.

Before using the table, it is important to determine if a binomial distribution should be used. In order to use this type of distribution, we must make sure that the following conditions are met:

1. We have a finite number of observations or trials.
2. The outcome of teach 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.

The binomial distribution gives the probability of r successes in an experiment with a total of n independent trials, each having probability of success p.   Probabilities are calculated by the formula C(n, r)pr(1 - p)n - r where C(n, r) is the formula for combinations.

Each entry in the table is arranged by the values of p and of r.  There is a different table for each value of n.

Other Tables

For other binomial distribution tables: n = 7 to 9, n = 10 to 11.  For situations in which np and n(1 - p) are greater than or equal to 10, we can use the normal approximation to the binomial distribution.  In this case, the approximation is very good and does not require the calculation of binomial coefficients.  This provides a great advantage because these binomial calculations can be quite involved.

Example

To see how to use the table, we will consider the following example from genetics.  Suppose that we are interested in studying the offspring of two parents who we know both have a recessive and dominant gene.  The probability that an offspring will inherit two copies of the recessive gene (and hence have the recessive trait) is 1/4.

Suppose we want to consider the probability that a certain number of children in a six-member family possesses this trait.  Let X be the number of children with this trait.  We look at the table for n = 6 and the column with p = 0.25, and see the following:

0.178, 0.356, 0.297, 0.132, 0.033, 0.004, 0.000

This means for our example that

• P(X = 0) = 17.8%, which is the probability that none of the children has the recessive trait.
• P(X = 1) = 35.6%, which is the probability that one of the children has the recessive trait.
• P(X = 2) = 29.7%, which is the probability that two of the children have the recessive trait.
• P(X = 3) = 13.2%, which is the probability that three of the children have the recessive trait.
• P(X = 4) = 3.3%, which is the probability that four of the children have the recessive trait.
• P(X = 5) = 0.4%, which is the probability that five of the children have the recessive trait.

n = 2

n = 3

n = 4

n = 5

n = 6

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