Chapter 4—Probability Distributions
Binomial Distribution
A random variable is said to be distributed binomially when four conditions are met:
1) There are a fixed number of n trials.
2) The trials are independent.
3) Each trial has two outcomes, success or failure.
4) Each trial has the same probability of success.
Consider the following problem:
“Eric is playing basketball. Suppose he makes 70% of his free throws, that each shot is
independent, and that the probability of making one is the same for all shots. Eric gets to shoot
four free throws.”
Let X = the number that he makes out of his four attempts.
1. Write something about each of the four conditions and how it is satisfied in this situation.
1)
2)
3)
4)
He’s making four shots.
Each shot is independent.
He will either hit or miss.
Each trial has probability .7 of making the shot.
2. What is the sample space for X?
0
1
Would you say that all outcomes are equally likely?
2
3
4
No
3. What is the probability that Eric makes all four shots?
(.7)^4 = .2401
4. What is P(X=0)?
(.3)^4 = .0081
5. Why is the probability that Eric makes “exactly three” free throws, P(X=3), NOT this:
(.7)(.7)(.7)(.3), .1029?
…because there are other ways in which he can make exactly 3. …(.7)(.7)(.7)(.3) is the
probability of his making the first three shots and missing the fourth.
6. List all possible ways to make 3 shots (of the four). Let H represent “hit” and M represent
“miss.” There are four ways.
H
H
H
M
H
H
M
H
H
M
H
H
M
H
H
H
7. What is the probability of each one of these?
.1029
8. Since they are mutually exclusive (they can’t happen at the same time, meaning there is no
intersection to subtract), we can simply add the four together. Find P(X=3).
4(.1029) = .4116
9. Make a probability distribution.
X
0
P(X)
.0081
1
2
3
.0756
.2646
.4116
4
.2401
10. We would say that X is distributed as B(n, p), where B stands for binomial (in the same way
that N stands for normal), parameter “n” is the number of observations/trials, and parameter “p”
is the probability of success. Write it down for this problem.
B(4, .7)
x
n x
n C x  p  (1  p)
Binomial Formula
n
same as :    p x  (1  p ) n x
 x
Example: Use the binomial formula to find the probability of Eric’s making exactly three
free throws.
3
43
4 C 3  .7  (1  .7)
4  (.7) 3  (.3)1
.4116
*We can use a calculator…
… to create a probability distribution.
“2nd, Distribution, binompdf, n, p”
binompdf (n, p)
Ex: Create a probability distribution for our basketball problem.
binompdf (4, .7)
… to find “exactly” X successes. 𝑃(𝑋 = #)
binompdf (n, p, X)
Ex: Find 𝑃(𝑋 = 3).
binompdf (4, .7, 3)
.4116
… to find cumulative successes, adding probabilities from the left. 𝑃(𝑋 ≤ #)
binomcdf (n, p, X)
Ex: Find 𝑃(𝑋 ≤ 3).
binomcdf (4, .7, 3)
.7599
Ex: Find 𝑃(𝑎𝑡 𝑚𝑜𝑠𝑡 2).
which is 𝑃(𝑋 ≤ 2)
binomcdf (4, .7, 3)
.3483
Ex: Find 𝑃(𝑋 < 3).
which is 𝑃(𝑋 ≤ 2)
binomcdf (4, .7, 3)
.3483
… to find cumulative successes, adding probabilities from the right. 𝑃(𝑋 ≥ #)
Rewrite it as: 𝟏 − 𝑷(𝑿 ≤ "x-1") and use binomcdf (n, p, X)
Ex: Find 𝑃(𝑋 ≥ 3).
1 − 𝑃(𝑋 ≤ 2)
1 − binomcdf (4, .7, 2)
.6517
Ex: Find 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 2).
1 − 𝑃(𝑋 ≤ 1)
1 − binomcdf (4, .7, 1)
.9163
Ex: Find 𝑃(𝑋 > 2).
1 − 𝑃(𝑋 ≤ 2)
1 − binomcdf (4, .7, 2)
.6517
… to find cumulative probabilities, “in the middle of”/between the X outcomes, not
including lowest or highest X values. 𝑃(# < 𝑋 ≤ #)
binomcdf (n, p, #) — binomcdf (n, p, #)
Ex: Find 𝑃(1 < 𝑋 ≤ 3).
𝑃(𝑋 ≤ 3) − 𝑃(𝑋 ≤ 1)
binomcdf (4, .7, 3) — binomcdf (4, .7, 1)
.6762
Ex: Find 𝑃(4 > 𝑋 > 0).
𝑃(𝑋 ≤ 3) − 𝑃(𝑋 ≤ 0)
binomcdf (4, .7, 3) — binomcdf (4, .7, 0)
.yy
Ex: Find 𝑃(3 ≥ 𝑋 ≥ 2).
𝑃(𝑋 ≤ 3) − 𝑃(𝑋 ≤ 1)
binomcdf (4, .7, 3) — binomcdf (4, .7, 1)
.zz
Ex: Find 𝑃(0 < 𝑋 ≤ 2).
𝑃(𝑋 ≤ 2) − 𝑃(𝑋 ≤ 0)
binomcdf (4, .7, 3) — binomcdf (4, .7, 0)
.yy