Transcript Binomial Distribution

Binomial Distribution
What
the binomial distribution is
How
to recognise situations where the
binomial distribution applies
How
to find probabilities for a given binomial
distribution, by calculation and from tables
When to use the binomial
distribution

Independent variables
Pascal’s Triangle
n
(a+b)
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
nCr
5C0
1
5C1
5
5C2
10
5C3
10
5C4
5
5C5
1
10 ways to get to the 3rd position numbering each of the terms
from 0 to 5. this can also be calculated by using nCr button on
your calculator 5C2=10
Pascal’s Triangle
n
(a+b)
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
nCr
n!÷(c!x(n-c)!)
5C0
5!÷(0!x5!)
1
5C1
5!÷(1!x4!)
5
5C2
5!÷(2!x3!)
10
5C3
5!÷(3!x2!)
10
5C4
5!÷(4!x1!)
5
5C5
5!÷(5!x0!)
1
A coin is tossed 7 times. Find the
probability of getting exactly 3
heads.
We could do Pascal's triangle or we could calculate:
7C3 x (P(H))7
The probability of getting a head is ½
 n
nCr   
r
7
7C 3   
 3
 
7
7
  1
1
35
      35 7 
 0.27
2 128
 3  2 
TASK

Exercise A Page 61
Unequal Probabilities


A dice is rolled 5 times
What is the probability it will show 6
exactly 3 times?
P(6’)=5/6
P(6)=1/6
 5
   5C 3  10
 3
 5  1   5 
         P(3 sixes in 5 rolls)
 3  6   6 
3
2
Task / Homework

Exercise B Page 62
The Binomial distribution is all
about success and failure.
When to use the Binomial Distribution
–
–
A fixed number ofX trials
Only two outcomes
–
–
(true, false; heads tails; girl,boy; six, not six …..)
Each trial is independent
IF the random variable X has Binomial
distribution, then we write X ̴ B(n,p)
Sometimes you have to
use the Binomial Formula
 n  x ( n x )
P( X  x)    p  q
,
 x
where q  1  p
Eggs are packed in boxes of 12. The probability
that each egg is broken is 0.35
Find the probability in a random box of eggs:
there are 4 broken eggs
12
P ( X  4)   0.354  0.65(124 )  495 0.354  0.658
4
 0.235 to 3 significant figures
Task / homework

Exercise C Page 65
Eggs are packed in boxes of 12. The probability
that each egg is broken is 0.35
Find the probability in a random box of eggs:
There are less than 3 broken eggs
P ( X  3)  P( X  0)  P ( X  1)  P( X  2)
12
12
12
0
(12 )
1
(11)
  0.35  0.65   0.35  0.65   0.352  0.65(10 )
0
1
2
 11 0.005688 12 0.351  0.6511  66 0.1225 0.01346 0.0151
USING TABLES of the
Binomial distribution
An easier way to add up binomial
probabilities is to use the cumulative
binomial tables
Find the probability of getting 3 successes in 6 trials,
when n=6 and p=0.3
n=6
x
0
1
2
3
4
5
6
P=0.3
P(X=x)
0.1176
0.4202
0.7443
0.9295
0.9891
0.9993
1.000
n=6
x
0
1
2
3
4
5
6
P=0.3
P(X=x) 0.1176 0.4202 0.7443 0.9295 0.9891 0.9993 1.000
http://assets.cambridge.org/97805216/05397/excerpt/9780521605397_excerpt.pdf
The probability of getting 3 or fewer successes is found by adding:
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1176 + 0.3026 + 0.3241 +
0.1852 = 0.9295
The probability of getting 3 or fewer successes is found by adding:
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1176 + 0.3026 + 0.3241
+ 0.1852 = 0.9295
This is a cumulative probability.
Task / homework

Exercise D page 67
Mean variance and standard deviation

μ = Σx x P(X=x)=mean


This is the description of how to get the mean
of a discrete and random variable defined in
previous chapter.
The mean of a random variable whos
distribution is B(n,p) is given as:

μ =np
Mean,

variance & standard deviation
σ²=Σx² x P(X=x) - μ²


is the definition of variance, from the last
chapter of a discrete random variable.
The variance of a random variable
whose distribution is B(n,p)
σ²= np(1-p)
 σ=
np(1

p)
TASK / HOMEWORK



Exercise E
Mixed Questions
Test Your self