Unit 10 Probability 102 Homework

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A particle starts at the origin and moves to and from on a straight line. At any move it jumps either 1 unit to the right or 1 unit to the left each with probability $\frac12$ . All successive moves are independent. Given that the particle is at the origin at the completion of the 6th move, find the probability that it never occupied a position to the left of the origin during previous moves.


Let's calculate the probability directly. Rows are the moves, columns are the locations, entries are the probability of being in this location on this move without ever having moved to the left of $0$.$$\beginarrayccccccc&0&1&2&3&4&5&6\\\hline 1&&\frac12\\2&\frac14&&\frac14\\3&&\frac28&&\frac18\\4&\frac 216&&\frac316&&\frac116\\5&&\frac532&&\frac432&&\frac132\\6&\frac564&&\frac964&&\frac564&&\frac164 \endarray$$


So my problem here is that I'm not quite getting the picture where I'm supposed to start. What I understand from the problem statement is that I need to first, do the change the variables in function $p(\textbfx)$, and then integrate this function over an interval of thickness $\epsilon$ to get the probability mass?


which would mean that $\rho(r)\epsilon$ is the probability mass (probability mass = probability density $\times$ volume ) of shell of radius $r$ and thickness $\epsilon=dr$, which is what I'm supposed to show. So I do the integral:


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Consider a system of two Einstein solids, A and B, each containing 10 oscillators, sharing a total of 20 units of energy. Assume that the solids are weakly coupled, and that the total energy is fixed.





The problem comes down to distinguishability. If the units of energy were distinguishable, then indeed it would behave as $\frac12^20$. Unit $p$ being in oscillator $A_1$ with unit $q$ being in oscillator $A_2$ would be counted as distinct from unit $p$ being in oscillator $A_2$ with unit $q$ being in oscillator $A_1$. This is what happens when their probabilities are truly independent. However, with indistinguishable units of energy, the two states described above only count once, showing that the probabilities of the two units are not totally independent. Now, the question is really, how many ways can we distribute 20 units of energy across 10 oscillators compared to across 20 oscillators? As the number of units increases, the number of microstates is reduced more and more, deviating farther and farther from the assumption of independent probabilities. Because of these reductions, the number of microstates with 19 units in $A$ and 1 unit in $B$ decreases relatively more than, say, with 18 units in $A$ and 2 units in $B$. So, the probabilities wind up being dependent on how many units are already in a given region. This is basically the source of the exchange interaction for otherwise non-interacting bosons.


One short, in class quiz will be given after each unit of the course, during the time usually reserved for in-class group exercises.These are primarily designed to verify your mathematical and conceptual understanding. As such, they will be collected and we

Unit 10 Probability 102 Homework

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