fall 2014 fadwa odeh (lecture 1). probability & statistics
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Fall 2014
Fadwa ODEH(lecture 1)
Probability & Statistics
Tossing a pair of dice
For one die, the probability of any face coming up is the same, 1/6. Therefore, it is equally probable that any number from one to six will come up.
For two dice, what is the probability that the total will come up 2, 3, 4, etc up to 12?
To calculate the probability of a particular outcome, count the number of all possible results. Then count the number that give the desired outcome. The probability of the desired outcome is equal to the number that gives the desired outcome divided by the total number of outcomes. Hence, 1/6 for one die.
List all possible outcomes (36) for a pair of dice.
Total Combinations How Many
2 1+1 1
3 1+2, 2+1 2
4 1+3, 3+1, 2+2 3
5 1+4, 4+1, 2+3, 3+2 4
6 1+5, 5+1, 2+4, 4+2, 3+3 5
7 1+6, 6+1, 2+5, 5+2, 3+4, 4+3 6
8 2+6, 6+2, 3+5, 5+3, 4+4 5
9 3+6, 6+3, 4+5, 5+4 4
10 4+6, 6+4, 5+5 3
11 5+6, 6+5 2
12 6+6 1
Sum = 36
2.8 5.6 8.3 11 14 17 14 11 8.3 5.6 2.8 %
361
362
363
364
365
366
365
364
363
362
361
Prob.
12 11 10 9 8 7 6 5 4 3 2 Total
Dice
0
0.05
0.1
0.15
0.2
2 3 4 5 6 7 8 9 10 11 12
Number
Pro
bab
ility
• Each possible outcome is called a “microstate”.• The combination of all microstates that give the same number of spots is called a “macrostate”.• The macrostate that contains the most microstates is the most probable to occur.
Combining Probabilities
If a given outcome can be reached in two (or more) mutually exclusive ways whose probabilities are pA and pB, then the probability of that outcome is: pA + pB
This is the probability of having either A or B
If a given outcome represents the combination of two independent events, whose individual probabilities are pA and pB, then the probability of that outcome is: pA × pB
This is the probability of having both A and B
Examples
Paint two faces of a die red. When the die is thrown, what is the probability of a red face coming up?
31
61
61 p
Throw two normal dice. What is the probability of two sixes coming up?
361
61
61
)2( p
Let p the probability of success (or desired event or outcome which is here 1/6 for one die). And let q the probability of failure (or undesired event or outcome which is here 5/6 for one die)
p + q = 1, or q = 1 – p
When two dice are thrown, what is the probability of getting only one six?
Probability of the six on the first die and not the second is:
Probability of the six on the second die and not the first is the same, so:
365
65
61 pq
185
3610
2)1( pqp
Probability of no sixes coming up is:
The sum of all three probabilities is:
p(2) + p(1) + p(0) = 1
3625
65
65
)0( qqp
p(2) + p(1) + p(0) = 1
pp+(pq+pq)+qq = 1
p² + 2pq + q² =1
(p + q)² = 1
The exponent is the number of dice (or tries).
Is this general?
Three Dice Example
(p + q)³ = 1
p³ + 3p²q + 3pq² + q³ = 1
p(3) + p(2) + p(1) + p(0) = 1
It works! It must be general?!
(p + q)N = 1
Binomial Distribution
Probability of n successes in N attempts
(p + q)N = 1
where, q = 1 – p.
nNnqpnNn
NnP
)!(!!
)(
Thermodynamic Probability
The term with all the factorials in the previous equation is the number of microstates that will lead to the particular macrostate. It is called the “thermodynamic probability”, wn.
)!(!
!
nNn
Nwn
Microstates
The total number of microstates is:
nw
nP
w
)(y probabilit True
For a very large number of particles
maxw
Mean (Average) of Binomial Distribution
nnPnPp
p
qpnNn
NnP
nnPn
nNn
n
)()( :Notice
)!(!!
)(
where
)(
pNn
pNqppNn
qpp
pnPp
pn
nPp
pnnPn
NN
N
n
nn
11 )1()(
)()(
)()(
Standard Deviation (σ)
222
22222
222
2
22
)(
nn
nnnnnnnnnn
nnnPnn
nn
n
pNqpNppNpNn
qpNpNqpNpn
qppNp
pqpp
pp
pn
nPp
pnnPn
NN
NN
nn
1
))(1()(
)()(
)()(
2
212
12
222
Npq
NpqpNpNNpq
pNpNqpN
nn
222
22
222
)()(
)(
For a Binomial Distribution
Npq
n
Npq
pNn
Coins
Toss 6 coins. Probability of n heads: so total number N choose n from it, could be written as
6
6
2
1
)!6(!
!6)(
2
1
2
1
)!6(!
!6
)!(!
!)(
nnnP
nnqp
nNn
NnP
nnnNn
n
N
For Six Coins
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 1 2 3 4 5 6
Pro
bab
ilty
Successes
Binomial Distribution
For 100 Coins
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
Pro
bab
ilty
Successes
Binomial Distribution
For 1000 Coins
Binomial Distribution
0
0.005
0.01
0.015
0.02
0.025
0.030 60 120
180
240
300
360
420
480
540
600
660
720
780
840
900
960
Successes
Pro
bab
ilty
Multiple Outcomes
NN
N
N
NNN
Nw
ii
i
!
!
!!!
!
321
We want to calculate lnW
Stirling’s Approximation
iii
i iiii
iii
i
NNNNw
NNNNNNw
NNNNN
Nw
NNNNN
)ln(lnln
)ln(lnln
!ln!ln!ln!ln!
!lnln
ln!ln: largeFor
Number Expected
Toss 6 coins N times. Probability of n heads:
Number of times n heads is expected is:
n = N P(n)
6
6
2
1
)!6(!
!6)(
2
1
2
1
)!6(!
!6
)!(!
!)(
nnnP
nnqp
nNn
NnP
nnnNn
Example: compute the multiplicities of macrostatesin an elementary (quantum!) model of a paramagnet
We can view the paramagnet as N magnetic moments each of which can be in 2 states either pointing parallel or anti-parallel to some given axis (determined, e.g., by an applied magnetic field). These states are referred to as “up" and “down", respectively. The total magnetization M along the given axis of the paramagnet is then proportional to the difference N up-Ndown = 2Nup-N.
Evidently, the macrostate specied by M has multiplicity given by the number of ways of choosing Nup magnetic moments to be \up" out of a total of N magnetic moments. We have
So paramagnet is like tossing a coin
upN
N
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