Download Activity #29

Math 103 - Cooley
Statistics for Teachers
OCC
Activity #29 – The Binomial Distribution
California State Content Standard - Statistics, Data Analysis, and Probability
N/A
Bernoulli Trials
Repeated trials of an experiment are called Bernoulli trials if the following conditions are satisfied:
1) The experiment (each trial) has only two possible outcomes: success or failure. The possible
outcomes are exactly the same for each trial.
2) The probability of success remains the same for each trial. (We use p for the probability of
success and q = 1 – p for the probability of failure.
3) The trials are independent. (The outcome of one trial has no influence on later trials.)
4) We are interested in the total number of successes, not the order in which they occur.
There may be 0, 1, 2, 3,…, or n successes in n trials.
Number of Outcomes Containing a Specified Number of Successes
n
In n Bernoulli trials, the number of outcomes that contain exactly x successes equals the binomial coefficient   .
 x
Binomial Probability Formula
Let X denote the total number of successes in n Bernoulli trials with success probability p. Then the
probability distribution of the random variable X is given by
n
P( X  x)    p x q n  x ,
x = 0, 1, 2, …, n.
 x
The random variable X is called a binomial random variable and is said to have the binomial distribution
with parameters n and p.
 Example #1:
Roll a standard die 3 times. Let the random variable X represent the number of times a 5 is rolled.
(So, a success is when a 5 is rolled in a single trial.)
a) What is the probability of a success? Answer: P(success) = p =
b) What is the probability of a failure? Answer: P(failure) = q =
5
6
1
6
.
. (Since q = 1 – p = 1 –
1
6
=
5
6
)
c) What is the probability of rolling exactly one 5?
1
2
 3  1   5 
Answer: P(X = 1) =        3  (.1667)  (.6944)  .3473
1   6   6 
d) What is the probability of rolling exactly two 5’s?
2
1
3  1   5 
Answer: P(X = 2) =        3  (.0278)  (.8333)  .0695
 2  6   6 
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 Example #2:
A multiple choice quiz of four questions is given. Each question has five possible answers. Let the random
variable X represent the number of questions answered correctly. If a student guesses at all the answers, find the
binomial distribution of X.
Solution:
The random variable X takes on the values 0, 1, 2, 3, and 4, the possible number of successes in four trials.
The probability of each value occurring is computed by using binomial trials with:
P(success) = P(answering the question correctly)

p=
P(failure) = P(answering the question incorrectly)

q=
X
1
(Since q = 1 – p = 1 –
1
5
=
4
5
)
P(X)
 4  1   4 
P(X = 0) =       = 1  1  (.4096) = .4096
0  5  5 
1
3
 4  1   4 
P(X = 1) =       = 4  (.2)  (.512) = .4096
1   5   5 
0
0
1
5
4
5
4
2
 4  1   4 
P(X = 2) =       = 6  (.04)  (.64) = .1536
 2  5   5 
3
 4  1   4 
P(X = 3) =       = 4  (.008)  (.80) = .0256
3  5   5 
4
4
0
 4  1   4 
P(X = 4) =       = 1  (.0016)  1 = .0016
 4  5   5 
2
3
2
1
Notice that the probabilities in the binomial distribution sum to 1. Now, we can ask follow up questions:
a)
What is the probability that a student answers exactly 3 out of 4 questions correct?
Answer: P(X = 3) = .0256
b)
What is the probability that a student answers at least 3 out of 4 questions correct?
Answer: P(X ≥ 3) = P(X = 3) + P(X = 4 )= .0256 + .0016 = .0272
c)
What is the probability that a student answers at most 1 out of 4 questions correct?
Answer: P(X ≤ 1) = P(X = 0) + P(X = 1) = .4096 + .4096 = .8192
d)
What is the probability that a student answers at least 1 question correct?
Answer: P(X  1) = 1 – P(X = 0) = 1 – .4096 = .5904
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Math 103 - Cooley
Statistics for Teachers
OCC
Activity #29 – The Binomial Distribution
 Exercises:
1)
A coin is tossed six times. Find the probability of its landing on tails exactly three times. (Ans: .3125)
2)
A certain drug was developed, tested, and found to be effective 70% of the time. Find the
probability of successfully administering the drug to at least nine of ten patients. (Ans: .1493)
3)
Ken Griffey Jr. has a lifetime batting average of .305. (This is the probability of getting a hit). If
he batted 5 times in one game, what is the probability that he gets at least 3 hits?. (Ans: .1698)
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