PROBABILITY AND THE BINOMIAL
DISTRIBUTION
Physics 270
LAWS OF CHANCE
o
o
o
o
o
Experiment with one die
m – frequency of a given result (say rolling a 4)
m/n – relative frequency of this result
All throws are equally probable.
The throws are independent.
The measured
relative frequency
tends to a limit –
1/6.
LAWS OF CHANCE
BINOMIAL PROBABILITY


deals with the probability of several successive
decisions, each of which has two possible
outcomes
If an event has a probability, p, of happening, then
the probability of it happening twice is p2, and in
general pn for n successive trials. If we want to
know the probability of rolling a die three times and
getting two fours and one other number (in that
specific order) it becomes:
P (44n)  P (4)P (4)P (n)

     0.023
1
6
2 5
6
BINOMIAL PROBABILITY



However this is only sufficient for problems where
the order is specific. If order is not important in the
above example, then there are 3 ways that 2 rolls
of four and 1 other could occur:
110, 101, 011, where 1 represents a roll of four and
0 represents a non-four roll.
Since there are 3 ways of achieving the same goal,
the probability is 3 times that of before, or 6.9%.
BINOMIAL DISTRIBUTION
BINOMIAL DISTRIBUTION
BINOMIAL DISTRIBUTION
General Formula
P (m )
m m n m
 Cn a b
Cnm
n!

m !(n  m)!
where m  n
 completely specified by n and a
 expectation value: m  na
 standard deviation:  = (nab)1/2
BINOMIAL DISTRIBUTION
Example:
A baseball player’s batting average is 0.333.
• The probability for a hit: a = 0.333
• The probability for an out: b = 1 a = 0.667
Average number of hits the player gets in 100 at bats
• µ = na = (100)(0.333) = 33
The standard deviation for 100 at bats
  nab  (100)(0.333)(0.667)  4.7
So, we can expect,
33 ± 4.7 hits for every 100 at bats
POISSON DISTRIBUTION
If the probability p is small and the number of observations is
large the binomial probabilities are hard to calculate.
In this instance it is much easier to approximate the binomial
probabilities by poisson probabilities. The binomial distribution
approaches the poisson distribution for large n and small p.
p(m) 
m 
 e
m!
POISSON DISTRIBUTION
The Poisson distribution can be derived from an approximation
of the Binomial distribution.
n!
P (m) 
am b n m
m !(n  m)!
n!
n(n  1)

(n  m)!
b n m  (1  a)n m
(n  m  1)(n  m)!
 nm
(n  m)!
a 2 (n  m)(n  m  1)
 1  a(n  m) 

2!
an 

 1  an 
2!
2

 e an
POISSON DISTRIBUTION
Example: Radioactive Decay
n = 1020 atoms, half-life  = 1012 y = 5 × 1019 s
• The probability for a decay: 1/ = 2 × 1020/s
Average number of decays per second:
• µ = na = (1020 atoms)(2 × 1020/s) = 2/s
zero decays
What’s the probability of more
than 1in
in
one second?
20 e 2 e 2 20 e 2 21e 2
p( 1)p(0)
1 p(0)  p
(1)  1 
 59.4%
 0.135 or 13.5%
0!
1!
0!
1
GAUSSIAN DISTRIBUTION
o when there are a large number of events per
observation
o requires many observations
o resembles a normal distribution (It is continuous!)
o
o
 is the standard deviation
 is the mean observed number of events
1
 ( r   )2 /2 2
p( r ) 
e
 2
GAUSSIAN DISTRIBUTION
The probability of r being in the range r1 and r2 is
given by
1
p( r1  r  r2 ) 
 2

r2
r1
e
 ( r   )2 /2 2
which cannot be evaluated analytically, (it can be
looked up in a table).
If the limits are at +/ ∞, then it normalizes to 1.
dr
GAUSSIAN DISTRIBUTION
It is very unlikely (< 0.3%) that a
measurement taken at random
from a Gaussian data set will be
more than ± 3σ from the true
mean of the distribution.
GAUSSIAN DISTRIBUTION
● An example illustrating the small difference between the
two distributions under the above conditions:
Consider tossing a coin 10 000 times.
■ a(heads) = 0.5
■ n = 10 000
❒ mean number of heads = μ = na = 5000
❒ standard deviation σ = [na(1  a)]0.5 = 50