The experiment is equivalent to tossing 128 coins at the same time. The probability of a coin landing on heads (yellow) is 1/2. Therefore, if all the coins are fair, after one try, one-half of the coins will land on tails (red). So, only 128/2=64 coins are tossed the second time, and 1/2 of them will land on tails.
After repeating the experiment n times, the expected number of counters that land on yellow at the end of the trial is
![f(n)=128((1)/(2))^n](https://img.qammunity.org/2023/formulas/mathematics/college/9yfqap9q5wksd5hu7ys1f99cwa1ncxywzs.png)
Part B)
Set f(n)=1 and solve for n, as shown below,
![\begin{gathered} f(n)=1 \\ \Rightarrow128((1)/(2))^n=1 \\ \Rightarrow(1)/(2^n)=(1)/(128) \\ \Rightarrow2^n=128 \\ \Rightarrow n\log 2=\log 128 \\ \Rightarrow n=(\log 128)/(\log 2)=7 \\ \Rightarrow n=7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i0znwf843fsoqyhih20j0z8dd5deq9x660.png)
Thus, after 7 trials, exactly one counter will land on yellow.