![](https://cdn.mathpix.com/cropped/2024_05_06_579637707648b9944b26g-14.jpg?height=312&width=2041&top_left_y=324&top_left_x=0)

On the calculator keyboard, there are digits from 0 to 9 and symbols for two operations (see the image). Initially, the display shows the number 0. You can press any keys. The calculator performs operations in the sequence of key presses. If the operation symbol is pressed several times in a row, the calculator will only remember the last press.

a) The button with the multiplication sign is broken and does not work. A scatterbrained Scientist pressed several buttons in a random sequence. Which result of the resulting chain of operations is more likely - an even number or an odd number?

b) Solve the previous problem if the button with the multiplication sign is fixed.

![](https://cdn.mathpix.com/cropped/2024_05_06_579637707648b9944b26g-14.jpg?height=503&width=468&top_left_y=978&top_left_x=795)

#
a) Let $A$ be the event "the result is odd".

Suppose with probability $p$ the penultimate number is odd. Then the result will remain odd only if an even number is added. Therefore, the conditional probability of an odd result is $p \cdot 1 / 2$.

The penultimate number was even with probability $1-p$. The result will be odd only if the last addend is odd. Therefore, in this case, the conditional probability of event $A$ is $(1-p) \cdot 1 / 2$.

By adding the probabilities of these mutually exclusive events, we get: $\quad \mathrm{P}(A)=1 / 2$

b) We will say that the Scholar has performed $n$ steps if he has entered $n$ numbers and performed some arithmetic operations $n-1$ times between them. Let $p_{n}$ be the probability that after $n$ steps, the number on the calculator is odd. Then $p_{n+1}=1 / 4+1 / 4 p_{n}$ (see problem 65336). From this, it is clear that if $p_{n} \leq 1 / 2$, then $p_{n+1}1$.

## Answer

a) Equally likely; b) the probability of an even number is greater.