4-  [ Increase and Decrease. Function Analysis ]

Ilya Muromets meets the three-headed Zmey Gorynych. Every minute, Ilya cuts off one head of the Zmey. Let $x$ be the resilience of the Zmey ( $x>0$ ). The probability $p_{s}$ that $s$ new heads will grow in place of the cut head ( $s=0$,

1, 2), is $\frac{x^{s}}{1+x+x^{2}}$. During the first 10 minutes of the battle, Ilya recorded how many heads grew back in place of each one he cut off. The resulting vector is $K=(1,2,2,1,0,2,1,0,1,2)$. Find the value of the Zmey's resilience such that the probability of the vector $K$ is maximized.
Due to the mutual independence of head regrowth (the number of heads that grow back depends only on vitality, but not on previous events), the probability of vector $K$ is $p_{1} p_{2} p_{2} p_{1} p_{0} p_{2} p_{1} p_{0} p_{1} p_{2}=\frac{x^{12}}{\left(1+x+x^{2}\right)^{10}}=\mathrm{P}(x)$.

We need to find the point at which this function takes its maximum value on the ray $(0,+\infty)$.

$\mathrm{P}^{\prime}(x)=\frac{12 x^{11}\left(1+x+x^{2}\right)^{10}-10\left(1+x+x^{2}\right)^{9}(1+2 x)}{\left(1+x+x^{2}\right)^{20}}=-\frac{2 x^{11}\left(4 x^{2}-x-6\right)}{\left(1+x+x^{2}\right)^{11}}$.

By setting the derivative to zero, we get the only positive root $x_{0}=\frac{\sqrt{97}+1}{8} \approx 1.36$.

Since $\mathrm{P}(x) \geq 0, \mathrm{P}(0)=0$ and $\mathrm{P}(x) \rightarrow 0$

as $x \rightarrow \infty$, then $x_{0}$ is the only point of maximum.

## Answer

$$
\frac{\sqrt{97}+1}{8} \approx 1.36
$$

| Problem $\underline{65336}$ Topics: | [ Discrete distribution  [ Linear recurrence relations  [ Limit of a sequence, convergence | Difficulty: 4-  Class 10.11 |
| :---: | :---: | :---: |
|  | $\frac{\text { Conditional probability }}{\text { Parity and monotonicity }}$ |  |

On the calculator keyboard, there are digits from 0 to 9 and signs of two operations (see figure). Initially, the display shows the number 0. Any keys can be pressed. The calculator performs operations in the sequence of key presses. If an operation sign is pressed several times in a row, the calculator will remember only the last press. A scatterbrained Scholar pressed a very large number of buttons in a random sequence. Find the approximate probability that the result of the resulting chain of operations is an odd number?

![](https://cdn.mathpix.com/cropped/2024_05_06_15085dfe3c1d08e007efg-44.jpg?height=623&width=397&top_left_y=635&top_left_x=836)

## Solution

We will say that the Scholar has taken $n$ steps if he has entered $n$ numbers, and between them $n-1$ times performed some arithmetic operations. Let $p_{n}$ be the probability that after $n$ steps, the calculator will show an odd number, and express $p_{n+1}$ in terms of $p_{n}$.

If the last operation was multiplication, the result will be odd only if both factors are odd; the probability of this is $1 / 2 p_{n}$.

If the last operation was addition, the result will be odd if the last addend differs in parity from the penultimate one; the probability of this is $1 / 2$.

The last operation can be either multiplication or addition, and both of these options are equally likely.

Therefore, the formula for total probability gives:

$p_{n+1}=1 / 2 \cdot 1 / 2+1 / 2 \cdot 1 / 2 p_{n}=1 / 4+1 / 4 p_{n}$.

Rewriting this equality as $p_{n+1}-1 / 3=1 / 4\left(p_{n}-1 / 3\right)$, we see that when $n$ increases by 1, the difference $p_{n}-1 / 3$ decreases by a factor of 4. Therefore, $p_{n}$ is close to $1 / 3$ for large $n$.

## Answer

$$
\frac{1}{3}
$$