Binomial Distribution

I. Introduction

The binomial distribution is a discrete distribution that describes the behavior of a count variable X if the following conditions apply:

1: The number of observations n is fixed.
2: Each observation is independent from every other observation.
3: Each observation represents one of TWO outcomes ("success" or "failure").
4: The probability of "success" is the same for each outcome; in other words, P("success") is consant.

If these conditions are met, then X has a binomial distribution with parameters n and p, abbreviated B(n, p).

Notation: X ~ B (n, p)

Some examples of binomial distributions:

• You flip a coin 5 times and calculate the probability of getting a certain number of heads.
• This meets each of the four requirements: there are a fixed number of observations (5 flips); each coin toss is independent of the others; each coin toss has only two possible outcomes (heads or tails); and the probability of success (getting heads) is constant (always 0.5) for each flip.
• This would be abbreviated as X~B(5, 0.5), where X is the number of heads.
• You take a 10 question multiple choice test with 4 answers per question and calculate the probability of getting a certain number of correct answers.
• This meets each of the four requirements: there are a fixed number of observations (10 questions); each question is independent of the others; each question has only two possible outcomes (correct or incorrect); and the probability of success (getting a question correct) is constant (always 0.25).
• This would be abbreviated as X~B(10, 0.25), where X is the number of correct answers.

II. Binomial Formula

If $$X \sim B(n,p)$$, then the probability that there are $$k$$ successes in $$n$$ independent observations can be calculated using the following binomial formula:

Binomial Formula: $$P(X = k)= {n \choose k} p^k (1-p)^{n-k} \:\:where \:\: {n \choose k} = \frac{n!}{k!(n-k)!}$$

Note: The $${n \choose k}$$ part (also written as $$_nC_k$$ and spoken as "n choose k") tells you how many paths/combinations can get you to k.

For example, let's say you pass three lights on your drive to school, and you want to know the probability of hitting two red lights. In other words, you want to know P(X = 2), where X equals the number of red lights. From using logic or drawing a tree diagram, you know that you can hit red/red/green, red/green/red, or green/red/red. Thus, there are three ways for X to equal 2. If you had used the formula above, you would have seen that $${n \choose k}$$ = $${3 \choose 2}$$ = $$\frac{3!}{2!(3-2)!}$$ = 3. Both from logic and from the formula, you can see that there are three ways to hit two red lights.

III. Calculating the Mean, Variance, and Standard Deviation

 Mean Standard Deviation Variance $$\mu = np$$ $$\sigma = \sqrt{npq}$$ $$\sigma^2 = npq$$

IV. Calculator!

Instead of using the binomial formula, you can use your calculator to find probabilities.

For P(X = k)

Hit 2nd Vars →   Scroll to A:binompdf → Fill in (n, p, k)

For P(X ≤ k)

Hit 2nd Vars   Scroll to B:binomcdf → Fill in (n, p, k)

**IMPORTANT NOTE:

Everything must be entered in the form of "less than or equal to" (≤). For problems that are "greater than" (>), you still must enter it as "less than or equal to" (≤) and then subtract this from 1.

The logic behind this is as follows: P(x > 54) + P(x ≤ 54) = 1.  Therefore, 1 - P(x ≤ 54) = P(x > 54).

In addition, if the problem is simply "less than" (<) and not "less than or equal to" (≤), you have to enter the number below into your calculator.

The logic behind this is as follows: P(x < 47) is equal to the P(x ≤ 46) because everything equal to or below 46 is less than 47.

TIP: Draw a number line and check off or shade the numbers you want to include. Then, decide where the cut off is and you can easily convert the problem into a "less than or equal to" scenario.

V. Example

At a city intersection, 35% of the vehicles that go through the intersection are SUVs. If 10 vehicles go through the intersection, calculate:

a. Calculate the probability that exactly 4 vehicles are SUVs.

$$P(X = 4)= {10 \choose 4} 0.35^4 (1-0.35)^{10-4}$$

$$P(X = 4)= \frac {10!}{4!(10-4)!} 0.35^4 (1-0.35)^{10-4}$$

$$P(X = 4)= \frac {10!}{4!6!} (0.35)^4 (0.65)^{6}$$

$$P(X = 4)= (210) (0.35)^4 (0.65)^{6}$$

$$P(X = 4)= 0.238$$

You can confirm your final answer by plugging this into your calculator using the steps above. You should still get a probability of 0.238.

b. Calculate the probability that 6 or more vehicles are SUVs.

← 1 2 3 4 5 6 7 8 9 10 →

As this number line shows, P(x ≥ 6) = 1 - P(x ≤ 5)

Using your calculator, you find that P(x ≤ 5)= 0.905.

Hence, P(x ≥ 6) = 1 - 0.905 = 0.095

c. Calculate the expected number of SUVs that go through the intersection.

Remember that "expected" refers to the mean. Using the formula given above, we know that the mean of a binomial distribution is equal to $$(n)\times(p)$$.

In this case, n is equal to 10 and p is equal to 0.35. Therefore, $$(n)\times(p)$$ = $$(10)\times(.35)$$ = 3.5

d. Calculate the standard deviation for the number of SUVs that go through the intersection.

Using the formula given above, we know that the standard deviation of a binomial distribution is equal to $$\sqrt{npq}$$.

In this case, n is equal to 10, p is equal to 0.35, and q is equal to 0.65. Therefore, $$\sqrt{npq}$$ = $$\sqrt{(10)(.35)(.65)}$$ = 1.508.