Goodies

On-line Binomial Distribution Utility

On-line Histogram Maker

The Binomial Random Variable

In many experiments there are only two outcomes. For instance:

• Flip a coin.
• Take a penalty shot on goal.
• Test a randomly selected circuit to see whether it is defective.
• Roll a die and determine whether it is a 6 or not.
• Determine whether there was flooding this year at Laguna Beach.

We call such an experiment a Bernoulli trial, and refer to the two outcomes -- often arbitrarily -- as success and failure. We will be interested in what happens when we perform a sequence of independent Bernoulli trials. For our random variable X we will count the number of successes. For instance:

• Flip a coin 10 times; X = number of heads.
• Test 50 randomly selected circuits from an assembly line; X = number of defective circuits.
• Roll a die 100 times; X = number of sixes you throw.
• Provide a property in Laguna Beach with flood insurance for 20 years; X is the number of years, during the 20-year period, during which the property is flooded.

Q What about the shots on goal example?
A That depends. If the shooter is improving (or getting worse), then the probability of success is changing, and X is not a binomial random variable. For X to be a binomial random variable, we require that the probability of success for each trial be fixed and independent of what happened before.

Q OK. Now I know what a binomial random variable is. How do we calculate its probability distribution?
A Calculating the probability distribution of X means calculating the following probabilities:

The probability of 1 success; P(x = 1)
The probability of 2 successes; P(x = 2),
and so on. The formula below is derived in the textbook. To use it, you should know how to compute the binomial coefficients C(m, n). Press here for a pop-up window that will help you compute them.

The Binomial Probability Distribution & Further Calculations

Now that we know how to calculate individual probabilities of the form P(X = x), let us put them all together to obtain a probability distribution:

Example
Your name is Edison and you are an expert penalty goal shooter. your skill has been improved for the past 10 years, and now you are as good as you will ever be. Your success rate has been measured at 80%. Thus, p = .8 and q = .2. You take n = 6 shots on goal, so the possible values of X (the number of successes) are 0, 1, 2, 3, 4, 5, 6. Here is the probability for each value of X:

P(X = 0) = C(6, 0) (.8)⁰(.2)⁶ = 1 1 .000064 = .000064
P(X = 1) = C(6, 1) (.8)¹(.2)⁵ = 6 .8 .00032 = .001536
P(X = 2) = C(6, 2) (.8)²(.2)⁴ = 15 .64 .0016 = .01536
P(X = 3) = C(6, 3) (.8)³(.2)³ = 20 .512 .008 = .08192
P(X = 4) = C(6, 4) (.8)⁴(.2)² = 15 .4096 .04 = .24576
P(X = 5) = C(6, 5) (.8)⁵(.2)¹ = 6 .32768 .2 = .393216
P(X = 6) = C(6, 6) (.8)⁶(.2)⁰ = 6 .262144 1 = .262144

We can use the probability distribution to find the probability that X is in a given range by adding the individual probabilities. (You can also use the On-line Binomial Distribution Utility to compute them.) For instance:

Probability of at least 5 successes	=	P(5 X 6)
	=	P(X = 5) + P(X = 6)
	=	.393216 + .262144 = .65536 (or a 65.5% chance)
Probability of at most 2 successes	=	P(0 X 2)
	=	P(X = 0) + P(X = 1) + P(X = 2)
	=	.000064 + .001536 + .01536 = .01696 (or a 1.7% chance)
Probability of at least 3 successes	=	P(3 X 6)
	=	1 - Probability of at most 2 successes
Since "at most 2 successes" and "at least 3 successes" are complementary events.
	=	1 - .01696 = .98304
We used the previous answer, and that was a easier than adding P(X = 3), P(X = 4), P(X = 5), and P(X = 6)

Now try some of the exercises in Section 8.2 of Finite Mathematics and Finite Mathematics and Applied Calculus