The probability the first bag still contains a red ball after moving several balls from bag to bag

combinatorics

Given a problem as follows:

There are $3$ distinct bags ($A, B$ and $C$).

  • The bag $A$ contains $1$ red ball, $2$ green balls and $3$ blue balls.
  • The bag $B$ contains $0$ red ball, $4$ green balls and $3$ blue balls.
  • The bag $C$ contains $0$ red ball, $4$ green balls and $2$ blue balls.

A person (with closed eyes) does the following actions in order:

  • Move $3$ balls from $A$ to $B$.
  • Move $3$ balls from $B$ to $C$.
  • Move $2$ balls from $C$ to $A$.

Randomness is guaranteed. Find the probability that the bag $A$ still contains a red ball.

Attempt

I think it is simple but I am not sure whether my logic is correct or not.
The subsequent actions depend on the previous actions.
The required probability is the same as the probability that the first action does not take a red balls and ignoring the remaining actions. More precisely, the probability that no red balls are moved in the last $2$ actions is $1$.

So my answer is

\begin{align}
\text{the required probability} &= \frac{{5 \choose 3}}{{6 \choose 3}}\times 1 \times 1\\
&= \frac{1}{2}
\end{align}

Is my reasoning correct? I am afraid of wrongly teaching my students.

Best Answer

As the commenters have mentioned, there are two possibilities for the movement of the red ball:

  1. The red ball does not move (probability $\frac 12$).
  2. The red ball moves from A to B to C back to A.

The probability of the latter case is $\frac 36 \cdot \frac{3}{7 + 3} \cdot \frac{2}{6 + 3} = \frac{1}{30}$, giving a total probability of $\frac{16}{30}$ that bag A contains a red ball.

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