Bug Trap Probability 9Ef5C4

1. **Problem statement:** A bug starts at point $B$ and moves randomly along the paths shown in the figure until it reaches one of the traps at points $A$, $F$, or $C$. We want to find the probability $p$ that the bug gets trapped at $F$, then compute $100p$. 2. **Understanding the problem:** The bug moves randomly along edges connecting points on the circle and inside it. The traps are at $A$, $F$, and $C$. Once the bug reaches any of these points, it stops. 3. **Assign variables:** Let $P_X$ be the probability that starting from point $X$, the bug eventually gets trapped at $F$. 4. **Known probabilities:** Since $A$, $F$, and $C$ are traps, $$P_A=0, \quad P_F=1, \quad P_C=0.$$ 5. **Points and connections:** - $B$ connects to $A$ and $C$. - $D$ connects to $A$, $B$, $E$, and $F$. - $E$ connects to $C$, $D$, and $F$. - $F$ connects to $D$, $E$, and is a trap. 6. **Write equations for $P_X$ at non-trap points:** The bug moves randomly along edges from $X$, so $$P_X = \frac{1}{\text{deg}(X)} \sum_{Y \sim X} P_Y,$$ where $\text{deg}(X)$ is the number of neighbors of $X$, and $Y \sim X$ means $Y$ is connected to $X$. 7. **Degrees and neighbors:** - $B$: neighbors $A$, $C$; degree 2 - $D$: neighbors $A$, $B$, $E$, $F$; degree 4 - $E$: neighbors $C$, $D$, $F$; degree 3 8. **Equations:** - For $B$: $$P_B = \frac{1}{2}(P_A + P_C) = \frac{1}{2}(0 + 0) = 0.$$ - For $D$: $$P_D = \frac{1}{4}(P_A + P_B + P_E + P_F) = \frac{1}{4}(0 + P_B + P_E + 1) = \frac{1}{4}(P_B + P_E + 1).$$ - For $E$: $$P_E = \frac{1}{3}(P_C + P_D + P_F) = \frac{1}{3}(0 + P_D + 1) = \frac{1}{3}(P_D + 1).$$ 9. **Substitute $P_B=0$ into $P_D$ equation:** $$P_D = \frac{1}{4}(0 + P_E + 1) = \frac{1}{4}(P_E + 1).$$ 10. **Substitute $P_D$ into $P_E$ equation:** $$P_E = \frac{1}{3}\left(\frac{1}{4}(P_E + 1) + 1\right) = \frac{1}{3}\left(\frac{P_E + 1}{4} + 1\right) = \frac{1}{3}\left(\frac{P_E + 1 + 4}{4}\right) = \frac{1}{3} \cdot \frac{P_E + 5}{4} = \frac{P_E + 5}{12}.$$ 11. **Solve for $P_E$:** $$P_E = \frac{P_E + 5}{12} \implies 12P_E = P_E + 5 \implies 11P_E = 5 \implies P_E = \frac{5}{11}.$$ 12. **Find $P_D$:** $$P_D = \frac{1}{4}\left(\frac{5}{11} + 1\right) = \frac{1}{4} \cdot \frac{16}{11} = \frac{16}{44} = \frac{4}{11}.$$ 13. **Recall $P_B=0$:** The bug starts at $B$, so the probability it gets trapped at $F$ starting from $B$ is $P_B=0$. 14. **Answer:** $$100p = 100 \times 0 = 0.$$