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The Binomial distribution
Cécile Ané
Stat 371
Spring 2006
Outline
(^1) Examples
(^2) General case
(^3) Assumptions
(^4) Mean and Standard deviation
Coin tossing example
Consider tossing a fair coin 3 times independently. Possible outcomes:
HHH HHT HTH HTT THH THT TTH TTT
Let Y = # of Heads.
IP{ Y = 3 } =
IP{ Y = 2 } = 3 (
Other example: experiment to mutate a gene in bacteria. The experiment was repeated 3 times independently with 3 colonies. Heads = got a mutation Tail = failure, no mutation.
Tossing an unfair coin
Same setting, but p = IP{Heads} = 0 .5. Mutagenesis experiment: p = IP{getting a mutation}.
HHH HHT HTH HTT THH THT TTH TTT
IP{ Y = 3 } = p^3 IP{ Y = 2 } = p p ( 1 − p ) + p ( 1 − p ) p + ( 1 − p ) p p = 3 p^2 ( 1 − p )
How do we get IP{ Y = j }, j = 1 , 2 , 3 without counting all possibilities?
General case
We have n coin tosses , with probability p of Heads. We conduct the trials independently. What is the probability of getting exactly j Heads?
IP{ Y = j } = n! j !( n − j )!
p j^ ( 1 − p ) n − j
= (^) n C (^) j p j^ ( 1 − p ) n − j
with
p j^ =
j times ︷ ︸︸ ︷ p p... p , p^0 = 1, and factorial notation: j! = j ( j − 1 )... 2 .1 and 0! = 1 Ex: 4! = 4. 3. 2. 1 = 24. cf. Table 2 p.674 for the (^) n C (^) j numbers.
General case
IP{ Y = j } =
n! j !( n − j )!
p j^ ( 1 − p ) n − j
Examples: p = .5 (fair coin), n = 3 tosses, j = 2 Heads. We get
IP{ Y = 2 } =
(. 5 )^2 (. 5 )^1 =
(. 5 )^3 = 3 / 8
as before. In general, for any p , we get
IP{ Y = 2 } =
p^2 ( 1 − p )^1 = 3 p^2 ( 1 − p )
as before.
Calculation examples
2 successes out of 7 trials, probability of success is p = .6: n = 7, j = 2
IP{ Y = 2 } =
(. 6 )^2 (. 4 )^5 =
(. 6 )^2 (. 4 )^5
= 21 (. 6 )^2 (. 4 )^5 =. 0774
5 failures out of 7 trials, probability of failure is p = .4: n = 7, j = 5
IP{ Y = 5 } =
(. 4 )^5 (. 6 )^2 =
(. 4 )^5 (. 6 )^2
= 21 (. 4 )^5 (. 6 )^2 =. 0774
same result...
Calculation examples
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 }
= 1 −
( 1 / 6 )^0 ( 5 / 6 )^6 = 1 − ( 5 / 6 )^6
Notation
Consider a random variable Y where Y = # of successes. Suppose we have n trials. We write
Y ∼ B( n , p )
B for binomial. In the last example we had Y ∼ B( 6 , 1 / 6 ). This notation is a shorthand for this distribution table:
y 0 1 2 3 4 5 6 IP{ Y = y } 0.335 0.402 0.200 0.054 0.008 0.0006 0.
describes Y ’s probability distribution, i.e. its probability mass function.
0.0 0 2 4 6 8 10
0.^
0.^
n= 10 , p= 0.
Probability
0.00 0 2 4 6 8 10
n= 50 , p= 0.
0.00 0 5 10 15 20 25
n= 200 , p= 0.
0.00 0 5 10 15 20
n= 20 , p= 0.
Probability
0 5 10 15 20
n= 20 , p= 0.
Some Possible Values
0.00 0 5 10 15 20
n= 20 , p= 0.
Underlying assumptions
Trials have exactly 2 outcomes The probability of success p is the same for all trials. Number of trials n is fixed in advance. If new patients are enrolled until one of them at least gets improved, then the binomial is not the correct distribution for Y. All trials are independent.
Underlying assumptions
Are assumptions met for Y = # kids with a cold, out of 20 in Mrs. Smith’s kindergarten class Y = # days with rain next week Y = # of students who answered “7” to the last question on the survey (pick a random number) Consider April 1, May 1, June 1,... , October 1. (7 dates) Y = # days with rain out of these 7 days. Drug trial on rats housed 3 in a cage
The binomial is a model. It is not reality. It is a way to provide structure on real world phenomena.
