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6.2 Homework Questions
Section 6.3
Binomial Random Variables
Binomial Setting

The four conditions for a binomial setting are:
Success/Failure
2. Independent Trials
3. Constant “p” (probability of success)
4. Set number of trials, n
1.
Geometric

1.
2.
3.
4.
The four conditions for a geometric
setting are:
Success/Failure
Independent Trials
Constant “p” (probability of success)
No set number of trials, n
Binomial Random Variable

The count of X successes in a binomial
setting is a binomial random variable.

The probability distribution of X is a
binomial distribution with parameters n
and p.

The possible values of X are the whole
numbers from 0 to n.
Binomial?
Genetics says that children receive genes from
each of their parents independently. Each child of
a particular pair of parents has probability 0.25 of
having type O blood. Suppose these parents have
5 children. Let X = the number of children with
type O blood.
 Shuffle a deck of cards. Turn over the first 10
cards, one at a time. Let Y = the number of aces
you observe.
 Shuffle a deck of cards. Turn over the top card.
Put the card back in the deck, and shuffle again.
Repeat this process until you get an ace. Let W =
the number of cards required.

Binomial Probabilities


Let’s do the children’s gene problem.
n=5 , p= 0.25
or
B(5, 0.25)

P(X=0)
P(none of the children have type O)=

P(X=1)
P(one child has type O)

P(X=2)
Building the formula
P(X = k) =
= P(exactly k successes in “n” trials)=
= (number of arrangements)∙ 𝑝𝑘 (1 − 𝑝)𝑛−𝑘
So we need a nice way of finding the
“number of arrangements”
Number of arrangements:
Binomial Coefficient


The number of ways of arranging k successes
among n observations is given by the
binomial coefficient:
𝑛
𝑛!
=
𝑘! 𝑛−𝑘 !
𝑘

You may know this as nCr

𝑛
CAUTION :
is NOT the fraction
𝑘
𝑛
𝑘
Binomial Probability

For B(n,p) , that is to say, for any Binomial:
𝑛 𝑘
P X=k =
𝑝 (1 − 𝑝)𝑛−𝑘
𝑘

This is on the formula sheet!
Examples:

Find the probability that exactly 3
children have type O blood.
B(5,0.25)

Should the parents be surprised if more
than 3 of their children have type O
blood? Justify your answer.
Mean and Standard Deviation of a
Binomial Distribution

Blood Type Probability Distribution:
X
0
1
2
3
4
5
P(X)
0.23730
0.39551
0.26367
0.08789
0.01465
0.00098

Above is the binomial from the blood
type problem.

Find the expected value and standard
deviation.
Mean and Standard Deviation of
Binomial Random Variables
Given B(n,p):
 𝜇𝑥 = 𝑛𝑝
 𝜎𝑥
= 𝑛𝑝(1 − 𝑝)
 Remember – these formulas ONLY work
for binomial distributions!

Both of these are on the formula sheet as
well.
Examples Continued:

Together, let’s do numbers Page 403: 69-72
Using the first one as an example B(20, 0.85) find:
𝑃(𝑋 = 17)

𝑃 𝑋 ≤ 12
𝑃 𝑋 = 0 + 𝑃 𝑋 = 1 + ⋯ + 𝑃(𝑋 = 12)
Homework (same as before)

Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92,
94-105)
Warm Up
Normal Approximation for Binomial
Distributions
As a rule of thumb, we will use the
Normal approximation when n is so
large that:
 𝑛𝑝 ≥ 10
 𝑛(1 − 𝑝) ≥ 10
 That is, the expected number of successes
and failures are both at least 10.


𝑛≤
1
𝑁
10
(independence)
Example:
Suppose that exactly 60% of all adult US
residents would say “agree” if asked if they
think shopping is frustrating. A survey asked
nationwide sampled 2500 adults.
 Let X = the number of people who agree.

◦ Show that X is approximately a binomial
random variable.
Check the conditions for using a Normal
approximation in this setting.
Example Continued:

Use a Normal distribution to estimate the
probability that 1520 or more of the
sample agree.
Find the mean
 Find the standard deviation
 Use the Normal curve

Homework #3 (again)

Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92, 94-105)