The Binomial Distribution

1
The Binomial Distribution

• Arrangements
• Remember the binomial theorem?

2
Expanding using arrangements
Arrangements of 4 As
(ab)4
aaaa
Arrangements of 3 As and 1 B
aaab aaba abaa baaa
Arrangements of 2 As and 2 Bs
aabb abab abba baab baba bbaa
abbb babb bbab bbba
Arrangements of 1 A and 3 Bs
bbbb
Arrangements of 4 Bs
The 1st line contains terms corresponding to a4
so coefficient is
The 2nd line contains terms corresponding to a3b
so coefficient is
The 3rd line contains terms corresponding to a2b2
so coefficient is
The 4th line contains terms corresponding to ab3
so coefficient is
The 5th line contains terms corresponding to b4
so coefficient is
3
Arrangements
A A A B B B B B B
How many ways are there of arranging these?
n 9
r 3
4
Example using a calculator
A A A B B B B B B
How many ways are there of arranging these?
n 9
r 3
To calculate this, type 9 followed by nCr
followed by 3 and press equals?
Use your calculator to work out
Explain your answer.
5
Binomial Theorem
A general rule for any expansion is
A special case occurs when a1 and bx
6
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

WRITTEN
which means the discrete random variable X is
binomially distributed
7
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

8
Binomial distribution

• Consider 3 trials
• n3
• 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(X0)
• Can happen one way defect-defect-defect
• (1-p)(1-p)(1-p)
• (1-p)3

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Binomial distribution

• If we want P(X1)
• 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

10
Binomial distribution
1 - good bulb 0 - defect bulb

• If we want P(X2)
• Can happen three ways 110, 011, 101
• pp(1-p)(1-p)ppp(1-p)p
• 3p2(1-p)

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Binomial distribution
1 - good bulb 0 - defect bulb

• We want P(X3)
• Can happen one way 111
• ppp
• p3

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Binomial distribution
P(X0) (1-p)3
P(X1) 3p(1-p)2
P(X2) 3p2(1-p)
P(X3) p3
r is the number of good bulbs
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Binomial distribution function

• In general, the binomial distribution function is
  given by

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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?

n 10
p 0.7
r 8
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Binomial distribution - example
n 10
p 0.9
r 8
0.233 (3 d.p.)
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Binomial distribution

• Typical shape of binomial
• Symmetric
• Mean and Mode approx pn

P
r
17
Binomial distribution - expected value

• For the binomial distribution

The mean value of X is given by np this is
also the expected value of X - E(X)
18
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