Download the binomial distribution - Ashburton College Maths Department

THE BINOMIAL DISTRIBUTION
Achievement Standard: 90646 (2.6) (part) external; credits 4
Key words: Trials, success/failure,
1. Definition of the Binomial distribution.
The Binomial Distribution occurs when:
(a) There is a fixed number (n) of trials.
(b) The result of any trial can be classified as a “success” or a “failure”
(c) The probability of a success ( or p) is constant from trial to trial.
n
 (d)
 Trials are independent.
x
 
If X represents the number of successes then:
n x
n-x
P(X =x) =    ( 1 -  )
x
2. Calculating Binomial probabilities.
Sigma p 317, Ex 15.01
Example 1
Calculate the probability of a drawing pin landing on its back 3 times in 5 trials
if the probability of getting a drawing pin to land on its back on any toss is 0.4
Answer: Let X be a random
 5  variable representing the number of “backs” in 5
 
tosses of the drawing
pin. Then X is binomial with n = 5,  = 0.4.
3
We want P(X = 3).
5
(i) By formula: P(X = 3) =   0.43 0.62
3
= 0.2304
(ii) Using tables: P(X = 3) = 0.2304
(iii) Graphics calculator: Menu, STAT, DIST, BINM, Bpd
Data Var
x: 3
Numtrial: 5
P: 0.4, execute
Note: Bpd gives individual binomial probabilities eg P(X=3),
Bcd gives cumulative binomial probabilities eg P(X ≤ 3)
Example 2
It is known that 60% of candidates will achieve in an examination.
If 5 people sit the examination find the probability:
(a) Exactly 3 candidates will achieve.
(b) At least 3 candidates will achieve.
Answer; Let X be a random variable representing the number of candidates
 5 (from

the 5) who achieve in the examination.
 Then

X is Binomial with n = 5,  = 0.6.
3
 
(i) Using the formula ( (a) only).
5
(a) P(X = 3) =  
3
0.63 0.42
= 0.3456.
[(ii) If you want to use tables we must redefine the problem:
Let Y represent the number of candidates who do not achieve. Then Y is
Binomial with n = 5,  = 0.4.
(a) P( X = 3) = P( 3 achieve)
= P (2 do not achieve)
= P ( Y = 2)
= 0.3456 (tables).
(b) P(At least 3 achieve) = P(X = 3, 4, 5)
= P(Y = 2, 1, 0)
= 0.3456 + 0. 2592 + 0.0778
= 0.6826]
Using your graphics calculator:
P (at least 3 achieve) = P( X ≥ 3)
= 1 – P( X ≤ 2)
= 1 - 0.31744 (using Bcd)
= 0.68256
Sigma p 322, Ex 15.02
Ex 15.03
3. Probalilities for combined events using Binomial probabilities
Example 4:
When administring a drug it was known that 86% of people using it were cured.
The testing programme administered the drug to two groups of 10 patients.
What is the probability that all 10 patients were cured in both groups?
answer: Let X be a random variable representing the number of patients
cured in a group of 10.
We can assume that X is binomial :
* a success is a cure, a failure is not a cure
* assume trial results from patient to patient are independent
* π = 0.86 is fixed
* n = 10 is fixed.
P( X = 10 ) = 0.2213 (graphics calculator)
P(X = 10) in both groups = 0.22132
= 0.04897
4. Use of sample statistics to estimate π for Binomial probabilities
Example 5:
Electrical components in computers fail independently of each other.
20 components were sampled and it was found that 3 of them failed.
Estimate the probability that, in a sample of 15 components, no more than 2 will
fail.
Answer: Let x represent the number of components that fail. Then we assume
that X is binomial with n = 15, π = 3 ÷ 20 = 0.15, (note that this is an
estimate of π).
P(X ≤ 2) = 0.60422 (graphics calculator)
5. The mean and variance of the Binomial Distribution.
The mean
 = n
The variance 2 = n(1-)
Proof: For one trial we have
So  = E[X],
=
X
P(X = x)
0
1-
1

E[X2] =  so Var[X] = E[X2] – 2
=  - 2
= (1 - )
Then for n independent trials:
E[X1+X2+…..Xn] = E[X1] + E[X2] +………+ E[Xn]
= n
Var[X1+X2+…..Xn] = Var[X1] + Var[X2] +………+ Var[Xn]
= n(1 - )
Example:
A drug is known to be 90% effective when it is used to cure a disease.
If 20 people are given the drug then X is binomial with n = 20,  = 0.9.
(a)  = n
= 20  0.9
= 18
(b) 2 = n(1-)
= 20  0.9  0.1
= 1.8
=
1·8
= 1.34 (2DP)
Sigma p 330, Ex 15.04