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