a) Every $5$ mins, a number will either go low ($-1$) or stay the same ($-0$). what is the chance of the number $4$ being $2$ after $20$ mins?
I can solve question "a" by simply drawing the question out. I want to know the equation behind the question.
b) Refer to Question "a)" If a person can add $+1$ at every $n$ minutes, what is the best interval to add $+1$ to maintain original number $4$ for longest?
For example) $+\frac{1}{5}$ mins, $+\frac{1}{10}$mins.
I will appreciate it with detail explanation.
Thank you
For a, you have four incidents. You need two of them to be -1 and two to be zero. Presumably (the question does not say) the chance of each is $\frac 12$, so you are asking the chance of two heads in four flips of a coin. Look up the binomial distribution.
b is badly stated. I would guess it means you start at 4, you do the 0/-1 thing every 5 minutes and get to add 1 however often you want. You then count how many times you have 4. Clearly on average you get a -1 every 10 minutes so you want to add 1 every 10 minutes.