# Binomial Table for n= 10 and n=11

## For n = 10 to n = 11

Of all discrete random variables, one of the most important due to its applications is a binomial random variable.  The binomial distribution, which gives the probabilities for the values of this type of variable, is completely determined by two parameters: and p.  Here n is the number of trials and p is the probability of success on that trial.  The tables below are for n = 10 and 11.  The probabilities in each are rounded to three decimal places.

We should always ask if a binomial distribution should be used.   In order to use a binomial distribution, we should check and see 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.

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 we have n = 2 to 6, n = 7 to 9. 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

The following example from genetics will illustrate how to use the table.  Suppose that we know the probability that an offspring will inherit two copies of a recessive gene (and hence end up with the recessive trait) is 1/4.

We want to calculate the probability that a certain number of children in a ten member family possesses this trait.  Let X be the number of children with this trait.  We look at the table for n = 10 and the column with p = 0.25, and see the following column:

.056, .188, .282, .250, .146, .058, .016, .003

This means for our example that

• P(X = 0) = 5.6%, which is the probability that none of the children has the recessive trait.
• P(X = 1) = 18.8%, which is the probability that one of the children has the recessive trait.
• P(X = 2) = 28.2%, which is the probability that two of the children have the recessive trait.
• P(X = 3) = 25.0%, which is the probability that three of the children have the recessive trait.
• P(X = 4) = 14.6%, which is the probability that four of the children have the recessive trait.
• P(X = 5) = 5.8%, which is the probability that five of the children have the recessive trait.
• P(X = 6) = 1.6%, which is the probability that six of the children have the recessive trait.
• P(X = 7) = 0.3%, which is the probability that seven of the children have the recessive trait.

n = 10

n = 11

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