Download The Binomial distribution

The Binomial distribution
Examples and Definition
Binomial Model (an “experiment”)
1
A series of n independent trials is conducted.
2
Each trial results in a binary outcome (one is labeled
“success’ the other “failure”).
3
The probability of success is equal to p for each trial,
regardless of the outcomes of the other trials.
Binomial Random Variable
The number of “successes” in the binomial experiment.
Let Y = # of success in the above model.
Then Y is a binomial random variable with parameters n
(sample size) and p (success probability). It is often
denoted
Y ∼ B(n, p).
Example: Tossing a fair coin
Toss a fair coin three times, P(H) = .5.
Interest is counting the number of heads.
Y = # of heads
“success” = heads; “failure” = tails
n=
p=
Y ∼
Example: Tossing a fair coin
Toss a fair coin three times, P(H) = .5.
Interest is counting the number of heads.
Y = # of heads
“success” = heads; “failure” = tails
n=3
p=
Y ∼
Example: Tossing a fair coin
Toss a fair coin three times, P(H) = .5.
Interest is counting the number of heads.
Y = # of heads
“success” = heads; “failure” = tails
n=3
p = .5
Y ∼
Example: Tossing a fair coin
Toss a fair coin three times, P(H) = .5.
Interest is counting the number of heads.
Y = # of heads
“success” = heads; “failure” = tails
n=3
p = .5
Y ∼ B(3, .5)
Example: Tossing an unfair coin
Toss a biased coin 5 times, P(H) = .7.
Interest is counting the number of heads.
Y = # of heads
“success” = heads; “failure” = tails
n=5
p = .7
Y ∼ B(5, .7)
Example: Counting Mutations
Experiment to mutate a gene in bacteria; the probability of
causing a mutation is .4. The experiment was repeated 10
times, with 10 independent colonies.
Interest is counting the number of mutations.
Y = # of mutations.
“success” = mutation; “failure” = no mutation
n = 10
p = .4
Y ∼ B(10, .4)
Computing Probabilities for a Binomial Random Variable.
Board Example
Tossing a biased coin; Y ∼ B(3, .7).
Tossing a fair Coin
Consider tossing a fair coin 3 times .
Y = # of heads in the 3 tosses; Y ∼ B(3, .5)
Consider the Possible outcomes:
TTT
HTT
THT
TTH
HHT
HTH
THH
HHH
1 1 1
1
. . =
2 2 2
8
1 1 1
3
IP{Y = 1} = 3
. .
=
2 2 2
8
1 1 1
3
IP{Y = 2} = 3
. . ) =
2 2 2
8
1
1 1 1
IP{Y = 3} =
. . =
2 2 2
8
IP{Y = 0} =
Tossing a Biased Coin
Consider tossing a biased coin 3 times; IP{H} = .7
Y = # of heads in the 3 tosses; Y ∼ B(3, .7)
Consider the Possible outcomes:
TTT
HTT
THT
TTH
HHT
HTH
THH
HHH
IP{Y = 0} = 1(.3 × .3 × .3) = 1(.70 )(.33 ) = .027
IP{Y = 1} =
IP{Y = 2} =
IP{Y = 3} =
Tossing a Biased Coin
Consider tossing a biased coin 3 times; IP{H} = .7
Y = # of heads in the 3 tosses; Y ∼ B(3, .7)
Consider the Possible outcomes:
TTT
HTT
THT
TTH
HHT
HTH
THH
HHH
IP{Y = 0} = 1(.3 × .3 × .3) = 1(.70 )(.33 ) = .027
IP{Y = 1} = 3 (.7 × .3 × .3) = 3(.71 )(.32 ) = .189
IP{Y = 2} =
IP{Y = 3} =
Tossing a Biased Coin
Consider tossing a biased coin 3 times; IP{H} = .7
Y = # of heads in the 3 tosses; Y ∼ B(3, .7)
Consider the Possible outcomes:
TTT
HTT
THT
TTH
HHT
HTH
THH
HHH
IP{Y = 0} = 1(.3 × .3 × .3) = 1(.70 )(.33 ) = .027
IP{Y = 1} = 3 (.7 × .3 × .3) = 3(.71 )(.32 ) = .189
IP{Y = 2} = 3 (.7 × .7 × .3) = 3(.72 )(.31 ) = .441
IP{Y = 3} =
Tossing a Biased Coin
Consider tossing a biased coin 3 times; IP{H} = .7
Y = # of heads in the 3 tosses; Y ∼ B(3, .7)
Consider the Possible outcomes:
TTT
HTT
THT
TTH
HHT
HTH
THH
HHH
IP{Y = 0} = 1(.3 × .3 × .3) = 1(.70 )(.33 ) = .027
IP{Y = 1} = 3 (.7 × .3 × .3) = 3(.71 )(.32 ) = .189
IP{Y = 2} = 3 (.7 × .7 × .3) = 3(.72 )(.31 ) = .441
IP{Y = 3} = 1(.7 × .7 × .7) = 1(.73 )(.30 ) = .343
Question
Is there a general formula for computing Binomial
probabilities?
We do not want to have to list all possibilities when n = 10,
or n = 100.
Background: Factorials
Factorial. Multiply all numbers from 1 to n (n is a positive
integer)
n! = n(n − 1)(n − 2)...(2)(1)
Example. n = 4.
n! = (4)(3)(2)(1) = 24.
Note. 0! = 1.
Background: Binomial Coefficients
n Cj
n Cj
=
n!
j!(n − j)!
n Cj
counts all of the ways that j successes can appear in n
total trials.
Example n = 5, j = 3.
(5)(4)
5!
=
= 10.
3!2!
2
cf. Table 2 p.674 for the n Cj numbers.
Chalkboard
Y ∼ B(n, p)
Can we “guess” the answer:
IP{Y = j} =?
Binomial Distribution Formula
IP{Y = j} =
n Cj
pj (1 − p)n−j
Y ∼ B(n, p) (a binomial random variable based on n trials
and success probability p)
The probability of getting j successes is given by
(j = 0, 1, ..., n)
IP{Y = j} =
=
pj (1 − p)n−j
n!
pj (1 − p)n−j
j!(n − j)!
n Cj
Example
IP{Y = j} =
n!
pj (1 − p)n−j
j!(n − j)!
Question:
You toss a fair coin 3 times, what is the probability of getting 2
heads.
Answer:
p = .5 (fair coin), n = 3 tosses, j = 2 Heads. We get
IP{Y = 2} =
3!
3.2.1
(.5)2 (.5)1 =
(.5)3 = 3/8
2! 1!
2.1 1
Example
IP{Y = j} =
n!
pj (1 − p)n−j
j!(n − j)!
Question:
Y ∼ B(7, .6). Compute IP{Y = 2}.
Answer:
p = .6, n = 7 tosses, j = 2 Heads. We get
7!
7.6 5!
(.6)2 (.4)5 =
(.6)2 (.4)5
2! 5!
1.2 5!
= 21(.6)2 (.4)5 = .0774
IP{Y = 2} =
Example
A new drug is available. Its success rate is 1/6: probability that
a patient is improved. I try it independently on 6 patients.
Probability that at least one patient improves?
p = 1/6, n = 6, j = 1, 2, 3, 4, 5 or 6.
IP{at least one improves} = IP{Y = 1 or Y = 2 or . . . or Y = 6}
= IP{Y = 1} + IP{Y = 2} + · · · + IP{Y = 6}
= 1 − IP{Y = 0}
6!
= 1−
(1/6)0 (5/6)6 = 1 − (5/6)6
0!6!
= .665
Probability Distribution
IP{Y = j} =
n!
pj (1 − p)n−j
j!(n − j)!
Y ∼ B(6, 1/6); n = 6, p = 1/6.
We can compute IP{Y = j}, for j = 0, 1, 2, 3, 4, 5, 6.
y
IP{Y = y }
0
0.335
1
0.402
2
0.200
3
0.054
4
0.008
5
0.0006
6
0.00002
Mean, Variance, and Standard Deviation for Binomial
Random Variables
Recall the Formula for Mean,Variance for a General
Discrete Random Variable
X
E(Y ) =
yi IP{Y = yi } ,
X
Var(Y ) =
(yi − µY )2 IP{Y = yi } ,
where the yi ’s are the values that the variable takes on and the
sum is taken over all possible values.
What if Y ∼ Bin(100, .5)
Do I have to sum over 101 different values?!?
Mean, Variance, and Standard deviation
If Y ∼ B(n, p) then
µ = IEY = np,
σ 2 = np(1 − p),
and
σ=
p
np(1 − p)
Example
Coin tossing: Y ∼ B(100, .5). Compute the mean, variance,
and standard deviation.
µY = np = 50
σY2 = np(1 − p) = 100(.5)(.5) = 25
√
σY =
25 = 5.
The assumptions for the binomial model
Underlying assumptions
A binomial random variable satisfies the following four
conditions, abbreviated BInS.
1
Binary outcomes. There are two possible outcomes for
each trial (success and failure).
2
Independent trials. The outcomes of the trials are
independent of each other.
3
n is fixed. The number of trials n is fixed in advance.
4
Same value of p. The probability of a success on a single
trial is the same for all trials.
Note
The binomial model with n trials is said to be made up of n
Bernoulli trials.
Bernoulli Trials
The Assumptions of Bernoulli Trials.
1
Each trial results in one of two possible outcomes, denoted
success (S ) or failure (F ).
2
The trials are independent.
3
The probability of S remains constant from trial-to-trial and
is denoted by p. Write q = 1 − p for the constant
probability of F .
Example - solve on the Chalk Board
IP{Y = j} =
n!
pj (1 − p)n−j
j!(n − j)!
Question: Devin Harris likes to play basketball. Assume that
Devin’s free throw attempts are Bernoulli trials with p = .8.
1
Devin will shoot four free throws. Calculate the probability
that he obtains S,S,F,S in that order
2
Devin will shoot four free throws. Calculate the probability
that he obtains a total of three successes.
3
Can you think of any reasons why the assumptions of the
binomial model might not be satisfied for free throw
shooting?
Examples for You
Example 3.45, p.106
Examples 3.47, 3.48 p.108-109
Example 3.49 p.109
Example 3.50 p.110 (when assumptions might be violated).