The Binomial Distribution ► Arrangements ► Remember the binomial theorem?
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Transcript of The Binomial Distribution ► Arrangements ► Remember the binomial theorem?
The Binomial Distribution
►Arrangements ►Remember the binomial
theorem?
Expanding using arrangements
(a+b)4 =
aaaa
+ aaab + aaba + abaa + baaa
+ aabb + abab + abba + baab + baba + bbaa+ abbb + babb + bbab + bbba
+ bbbb
Arrangements of 4 As
Arrangements of 3 As and 1 B
Arrangements of 2 As and 2 Bs
Arrangements of 1 A and 3 BsArrangements of 4
Bs
The 2nd line contains terms corresponding to a3b so coefficient isThe 3rd line contains terms corresponding to a2b2 so coefficient isThe 4th line contains terms corresponding to ab3 so coefficient is
The 1st line contains terms corresponding to a4 so coefficient is
The 5th line contains terms corresponding to b4 so coefficient is
10
4
41
4
62
4
43
4
14
4
Arrangements
How many ways are there of arranging these?
)123)(123456(
123456789
!3!6
!9
A A A B B B B B B
84123
789
)!(!
!
rnr
n
r
n
n = 9
r = 3
)!39(!3
!9
3
9
Example – using a calculator
How many ways are there of arranging these?
A A A B B B B B B
n = 9
r = 3
3
9
To calculate this, type “9” followed by “nCr” followed by “3” and press equals?
Use your calculator to work out
Explain your answer.
6
9
Binomial Theorem
nnnnnn bban
ban
ban
aba
33221
321)(
nn xxn
xn
xn
x
32
3211)1(
A general rule for any expansion is…
A special case occurs when a=1 and b=x …
The Binomial Distribution•The binomial distribution is a discrete distribution defined by 2 parameters
•the number of trials - n•the probability of a success - p
),(~ pnBXWRITTEN :
… which means the discrete random variable X is binomially distributed
Binomial distribution - example
►Testing for defects “with replacement” Have many light bulbs Pick one at random, test for defect, put it
back Pick one at random, test for defect, put it
back If there are many light bulbs, do not have
to replace
Binomial distribution►Consider 3 trials
n=3
►p is the probability of picking a good bulb so (1-p) is the probability of picking a defect bulb
►the random variable X is the measure of the number of good bulbs
►If we want P(X=0): Can happen one way: defect-defect-defect (1-p)(1-p)(1-p) (1-p)3
),(~ pnBX
Binomial distribution
►If we want P(X=1): Can happen three ways: 100, 010, 001 p(1-p)(1-p)+(1-p)p(1-p)+(1-p)(1-p)p 3p(1-p)2
For simplicity:1 - good bulb0 - defect bulb
Binomial distribution
►If we want P(X=2): Can happen three ways: 110, 011, 101 pp(1-p)+(1-p)pp+p(1-p)p 3p2(1-p)
1 - good bulb0 - defect bulb
Binomial distribution
►We want P(X=3): Can happen one way: 111 ppp p3
1 - good bulb0 - defect bulb
Binomial distributionP(X=0): (1-p)3
P(X=3): p3
P(X=1): 3p(1-p)2
P(X=2): 3p2(1-p)
rr pprXP 31 waysof #
r is the number of good bulbs
Binomial distribution function
In general, the binomial distribution function is given by:
rnr pprXP 1 waysof #
!!
!
rnr
nC rn
rnrr
n ppCrXP 1
rnrr
n qpCrXP or
q = 1 -p
Binomial distribution - example
►Testing for defects “with replacement” Suppose 10 bulbs were tested
The probability of a good bulb is 0.7
What is the probability of there being 8 good bulbs in the test?
rnrr
n ppCrXP 1
n = 10
p = 0.7
r = 8
Binomial distribution - example
rnrr
n ppCrXP 1
n = 10 p = 0.9 r = 8
81088
10 7.017.08 CXP
45!2!8
!10
!!
!8
10
rnr
nC
28 3.07.0458 XP
= 0.233 (3 d.p.)
Binomial distribution
►Typical shape of binomial: Symmetric Mean and Mode approx = p*n
r
P
Binomial distribution - expected value
►For the binomial distribution
),(~ pnBX
The mean value of X is given by np
… this is also the expected value of X - E(X)
Binomial distribution - example
►Testing for defects “with replacement” Suppose 10 bulbs were tested The probability of a good bulb is 0.7 What is the probability of there being 8
good bulbs in the test?
The mean (expected value)?
= 0.233 (3 d.p.)
The mean value of X is given by np= 10 x 0.7 = 7