Answer:
Following are the solution to the given question:
Explanation:
Please find the complete and correct question in the attachment file.
For Point a:
For the second round,
A is selects kA(2) either 0 or 1, so for each of them, that is
.
B selects
for each choice with the probability of
.
If
wins the second rear race.
![\to P[A \ wins] = P[kA(2) < kB(2)]](https://tex.z-dn.net/?f=%5Cto%20P%5BA%20%5C%20wins%5D%20%3D%20P%5BkA%282%29%20%3C%20kB%282%29%5D)
![= P[kA(2) = 0] \times P[kB(2) > 0] + P[kA(2) = 1] \times P[kB(2) > 1]\\\\= \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{2}{4} \\\\=\frac{3}{8} +\frac{2}{8} \\\\= \frac{3+2}{8}\\\\= \frac{5}{8}](https://tex.z-dn.net/?f=%3D%20P%5BkA%282%29%20%3D%200%5D%20%5Ctimes%20P%5BkB%282%29%20%3E%200%5D%20%2B%20P%5BkA%282%29%20%3D%201%5D%20%5Ctimes%20P%5BkB%282%29%20%3E%201%5D%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B3%7D%7B4%7D%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20%5Cfrac%7B2%7D%7B4%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B3%7D%7B8%7D%20%20%2B%5Cfrac%7B2%7D%7B8%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B3%2B2%7D%7B8%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B5%7D%7B8%7D)
For Point b:
Throughout this example,
also selects to be either 0 or 1 with such a
probability. So, although B chooses
from
the probabilities each are
:
![= P[kA(3) = 0] \times P[kB(3) > 0] + P[kA(3) = 1] \times P[kB(3) > 1]\\\\= \frac{1}{2} \times \frac{7}{8} + \frac{1}{2} \times \frac{6}{8}\\\\= \frac{7}{16} + \frac{6}{16}\\\\= \frac{7+6}{16} \\\\= \frac{13}{16}\\\\](https://tex.z-dn.net/?f=%3D%20P%5BkA%283%29%20%3D%200%5D%20%5Ctimes%20P%5BkB%283%29%20%3E%200%5D%20%2B%20P%5BkA%283%29%20%3D%201%5D%20%5Ctimes%20P%5BkB%283%29%20%3E%201%5D%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B7%7D%7B8%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B6%7D%7B8%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B7%7D%7B16%7D%20%2B%20%5Cfrac%7B6%7D%7B16%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B7%2B6%7D%7B16%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B13%7D%7B16%7D%5C%5C%5C%5C)
For point c:
Assume that B tries again 16 times (typical value), and it destroys. In addition, throughout the exponential background n is obtained at 10 when choosing k between 0 to 2n−1. The probability of A winning all 13 backoff events is: ![P[A \text{wins remaining races}] = 16\pi i =4P[A \ wins \ i |A \ wins \ i -1 ]](https://tex.z-dn.net/?f=P%5BA%20%5Ctext%7Bwins%20remaining%20races%7D%5D%20%3D%2016%5Cpi%20i%20%3D4P%5BA%20%5C%20wins%20%5C%20i%20%7CA%20%5C%20wins%20%5C%20i%20-1%20%5D)
Let the k value kA(i) be A for the backoff race I select. Because A retains the breed
![=(kA(i)] \cdot P[kA(i+ 1)< kB(i+ 1)] \geq P[kA(i) + 1](https://tex.z-dn.net/?f=%3D%28kA%28i%29%5D%20%5Ccdot%20P%5BkA%28i%2B%201%29%3C%20kB%28i%2B%201%29%5D%20%5Cgeq%20P%5BkA%28i%29%20%2B%201%3CkB%28i%29%5D%20%5Ccdot%20%20P%5BkA%28i%2B%201%29%3C%20kB%28i%2B1%29%5D%2BP%5BkA%28i%29%20%2B%201%20%5Cgeq%20kB%28i%29%5D%20%5Ccdot%20%20P%5BkA%28i%2B%201%29%3C%20kB%28i%2B%201%29%5D%20%5C%5C%5C%5C%3D%20%28P%5BkA%28i%29%20%2B%201%3C%20kB%28i%29%5D%20%2BP%5BkA%28i%29%20%2B%201%20%5Cgeq%20kB%28i%29%5D%29%20%5Ctimes%20P%5BkA%28i%2B%201%29%20%3C%20kB%28i%2B%201%29%5D%5C%5C%5C%5C%3DP%5BkA%28i%2B%201%29%3C%20kB%28i%2B%201%29%5D%5C%5C%5C%5C)
For point d:
Two stations A and B are supposed. They assume that B will try 16 times afterward. Even so, for A, 16 races were likely to also be won at a rate of 0.82 For Just higher expectations of three A, B, and C stations. For Station A, possibility to win all backoffs