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Fall 2014 Fadwa ODEH (lecture 1)
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Probability & Statistics
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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?
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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.
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List all possible outcomes (36) for a pair of dice. Total CombinationsHow Many 21+11 31+2, 2+12 41+3, 3+1, 2+23 51+4, 4+1, 2+3, 3+24 61+5, 5+1, 2+4, 4+2, 3+3 5 71+6, 6+1, 2+5, 5+2, 3+4, 4+36 82+6, 6+2, 3+5, 5+3, 4+45 93+6, 6+3, 4+5, 5+44 104+6, 6+4, 5+53 115+6, 6+52 126+61 Sum = 36
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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.
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Combining Probabilities If a given outcome can be reached in two (or more) mutually exclusive ways whose probabilities are p A and p B, then the probability of that outcome is: p A + p B 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 p A and p B, then the probability of that outcome is: p A × p B This is the probability of having both A and B
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Examples Paint two faces of a die red. When the die is thrown, what is the probability of a red face coming up? Throw two normal dice. What is the probability of two sixes coming up?
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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?
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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:
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Probability of no sixes coming up is: The sum of all three probabilities is: p(2) + p(1) + p(0) = 1
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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?
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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
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Binomial Distribution Probability of n successes in N attempts (p + q) N = 1 where, q = 1 – p.
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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”, w n.
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Microstates The total number of microstates is: For a very large number of particles
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Mean (Average) of Binomial Distribution
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Standard Deviation (σ)
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For a Binomial Distribution
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Coins Toss 6 coins. Probability of n heads: so total number N choose n from it, could be written as
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For Six Coins
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For 100 Coins
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For 1000 Coins
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Multiple Outcomes We want to calculate lnW
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Stirling’s Approximation
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Number Expected Toss 6 coins N times. Probability of n heads: Number of times n heads is expected is: n = N P(n)
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Example: compute the multiplicities of macrostates in 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 -N down = 2N up -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
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