probability of discrete random variable

Question

In a nuclear reaction, a particle can either separate into two pieces, or not separate, with respective probabilities $2/3$ and $1/3$. Knowing that pieces behave like new independent particles, find the law of probability and mean of the number of particles obtained after two reactions from of a single particle.

Let $X$ be the random variable which designates the number of pieces obtained after two reactions. It is clear that $X\in \{1, 2, 3, 4\}$ and the probability distribution of this variable ${(K, P (X = k)) / 1 ≤ k ≤ 4}$

And he came after that and give me the probability of each case.

$P(k=1) = 3/27$

$P(k=2) = 8/27$

$P(k=3) = 8/27$

$P(k=4) = 8/27$

Can anyone help me to prove this solution?

Answer 1

You start with only one particle, let's call it $A$. The set of your particles at the beginning is $$\{A\}.$$

After the first reaction, the particle $A$ can produce another particle $B$, or not. Then we can have the following particle sets:

$$\begin{cases} \{A\} & ~\text{with probability}~\frac{1}{3}\\ \{A,B\} & ~\text{with probability}~\frac{2}{3}\\ \end{cases}$$

After the second reaction, we can have various scenario:

1. we start from $\{A\}$ and we don't produce anything with probability $\frac{1}{3}$, obtaining $\{A\}$;
2. we start from $\{A\}$ and we produce a new particle $C$ with probability $\frac{2}{3}$, obtaining $\{A,C\}$.
3. we start from $\{A,B\}$ and $A$ produce a new particle $D$ with probability $\frac{2}{3}$, while $B$ produce a new particle $E$ with probability $\frac{2}{3}$ obtaining $\{A,B,D,E\}$ (total probability $\left(\frac{2}{3} \cdot \frac{2}{3}\right)$).
4. we start from $\{A,B\}$ and $A$ produce a new particle $F$ with probability $\frac{2}{3}$, while $B$ does not produce anything with probability $\frac{1}{3}$ obtaining $\{A,B,F\}$ (total probability $\left(\frac{2}{3} \cdot \frac{1}{3}\right)$).
5. we start from $\{A,B\}$ and $A$ does not produce anything with probability $\frac{1}{3}$, while $B$ produce the particle $G$ with probability $\frac{2}{3}$ obtaining $\{A,B,G\}$ (total probability $\left(\frac{1}{3} \cdot \frac{2}{3}\right)$).
6. we start from $\{A,B\}$ and $A$ does not produce anything with probability $\frac{1}{3}$, while $B$ does not produce anything with probability $\frac{1}{3}$ obtaining $\{A,B\}$ (total probability $\left(\frac{1}{3} \cdot \frac{1}{3}\right)$).

Summaryzing:

$$\begin{array}{cccc} \text{Beginning} & \text{After first reaction} & \text{After second reaction} & \text{Probability}\\ \{A\} & \{A\} & \{A\} & \frac{1}{3} \cdot \frac{1}{3} = \frac{3}{27}\\ \{A\} & \{A\} & \{A,C\} & \frac{1}{3} \cdot \frac{2}{3} = \frac{6}{27}\\ \{A\} & \{A,B\} & \{A,B,D,E\} & \frac{2}{3} \cdot \left(\frac{2}{3} \cdot \frac{2}{3}\right) = \frac{8}{27} \\ \{A\} & \{A,B\} & \{A,B,F\} & \frac{2}{3} \cdot \left(\frac{2}{3} \cdot \frac{1}{3}\right) = \frac{4}{27}\\ \{A\} & \{A,B\} & \{A,B,G\} & \frac{2}{3} \cdot \left(\frac{1}{3} \cdot \frac{2}{3}\right) = \frac{4}{27}\\ \{A\} & \{A,B\} & \{A,B\} & \frac{2}{3} \cdot \left(\frac{1}{3} \cdot \frac{1}{3}\right) = \frac{2}{27}\\ \end{array}.$$

Now, you have only one case where at the end you have only one particle. This happens with probability $$p(k=1)=\frac{3}{27}.$$

Further, you have two cases where at the end you have two particles ($\{A,C\}$ and $\{A,B\}$. Summing up, you get: $$p(k=2) = \frac{6}{27} + \frac{2}{27} = \frac{8}{27}.$$

You have two cases where at the end you have three particles ($\{A,B,F\}$ and $\{A,B,G\}$), Summing up: $$p(k=3) = \frac{4}{27} + \frac{4}{27} = \frac{8}{27}.$$

Finally, in only one case you have four particles ($\{A,B,D,E\}$), with probability $$p(k=4) = \frac{8}{27}.$$