Particle Physics

Solution

Answer: 20.5%

Let \(k\) be the number of \(+1\) steps. Since there are \(10\) steps total, the number of \(-1\) steps is \(10-k\). The final position is \[ (+1)\cdot k + (-1)\cdot (10-k). \] Simplifying, \[ k - (10-k) = 2k - 10. \] \[ k = 6. \] So the particle must take \(6\) steps of \(+1\) and \(4\) steps of \(-1\). The number of ways to arrange \(6\) positive steps among \(10\) total steps is \[ \binom{10}{6} = 210. \] Each specific sequence of \(10\) steps has probability \[ \left(\frac{1}{2}\right)^{10}. \] Therefore the probability of ending at position \(2\) is \[ \binom{10}{6}\left(\frac{1}{2}\right)^{10}. \] \[ \frac{105}{512} \approx 0.205. \] So the probability that the particle is at position \(2\) after \(10\) steps is approximately \(0.205\), or about \(20.5\%\).