The Absent-Minded Scientist in his laboratory derived a single-celled organism, which with a probability of 0.6 divides into two such organisms, and with a probability of 0.4 dies without leaving offspring. Find the probability that after some time the Absent-Minded Scientist will not have any such organisms left.
It doesn't matter how much time is spent. Therefore, for simplicity, we will assume that organisms either divide or die every second, but strictly one at a time. When something happens to one of them, the others patiently wait their turn. By making this assumption, we get a standard random walk problem: every second, the number of organisms either increases by one (with probability \( p = 0.6 \)) or decreases by one (with probability \( q = 0.4 \)).

The number of organisms at a certain moment will be called the state of the population. Let \( x \) be the probability that the number of organisms will decrease by 1 after several steps.

Initially, the population is in state 1, and we are interested in the transition probability \( 1 \rightarrow 0 \), which is precisely the value of \( x \).

The population can reach state 0 in two ways:

1) In the first second, the only existing organism dies. The probability of this is \( q \).
2) In the first second, the only organism divides, and the population transitions to state 2. The probability of this event is \( p \). The subsequent transition

\( 2 \rightarrow 0 \) has a probability of \( x^2 \), since it consists of two independent transitions \( 2 \rightarrow 1 \) and \( 1 \rightarrow 0 \), each with a probability of \( x \).

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-03.jpg?height=183&width=366&top_left_y=2295&top_left_x=846)

Thus, the law of total probability gives the equation \( x = q + p x^2 \).

From this, \( x = 1 \) or \( x = q / p \). We need to determine which root is extraneous. Consider \( x \) as a function of \( p \). Plotting the lines \( x(p) = 1 \) and \( x(p) = 1 - p / p \) in the coordinate system \( p \) O, we use the following considerations.

1) The function \( x(p) \) is continuous.
2) \( x(p) \leq 1 \).
3) \( x(1) = 0 \) - if organisms do not die but only divide, the population will never die out with certainty.

These three conditions are satisfied by the function \( x(p) = \left\{\begin{array}{l}1, \text{ if } 0 \leq p \leq \frac{1}{2}, \\ \frac{q}{p}, \text{ if } \frac{1}{2} < p \leq 1 \end{array}\right. \).

Since \( p = 0.6 > 1 / 2 \), we have \( x = q / p = 0.4 / 0.6 = 2 / 3 \).

## Answer

\( 2 / 3 \).