Solve the equation for max value of $LHS$ : $\binom{n-1}{4} \cdot (0.8)^5 \cdot (0.2)^{n-1-4} < 1$

I am stuck in the following problem and need some help.

The probability that an experiment has a successful outcome is 0.8. The experiment is to be repeated until five successful outcomes occurred. How many repetitions are required in order to have 5 successful outcomes?

The possible solution I could come up with is by solving the following equation for $n$. $$\binom{n-1}{4} \cdot (0.8)^5 \cdot (0.2)^{n-1-4} < 1$$ Now, how to find the value of $n$ such that LHS is max. Kindly comment.

• Hint:$$\binom{n-1}{4} \cdot (0.8)^5 \cdot (0.2)^{n-1-4} <1$$ Aug 26, 2017 at 5:13
• @Khosrotash How to find the value of $n$ such that LHS is max? Aug 26, 2017 at 5:26
• @AbhinavGupta. Just compute it for some small values of $n$. Notice that, using whole numbers, the lhs write $$\frac{128}{3} 5^{-n} (n-4) (n-3) (n-2) (n-1)$$ Aug 26, 2017 at 6:16

Your solution is correct and it is the probability of exactly 5 success in N throws. If you want to find the N which maximize the probability you could easily use derivate. As Claude Leibovici said in the comments: $$P(n)=\frac{128}{3}5^{-n}(n-4)(n-3)(n-2)(n-1)$$ If we simplify we get: $$P(n)=\frac{128}{3}5^{-n}(n^2-7n+12)(n^2-3n+2)=\frac{128}{3}5^{-n}(n^4-10n^3+35n^2-50n+24)$$ Then we should apply derivative: $$\frac{dP(n)}{dn}=0$$ Calculating this term is not very completed only needs a little patience. But after derivation the equation will not have a closed form, you will need numerical solution to solve it.
• It is easier to find the ratio $P(n)/P(n-1)$ and find when it crosses the value $1$. So long as it is above $1$, you are increasing; when it dips below 1, you are decreasing. Aug 26, 2017 at 7:45