To solve this, we will first assign probabilities based on the conditions and then calculate the required probability.

Step 1: Assign probabilities

Let the probability of student $C$ winning be $p$. According to the problem:

1. $A$ and $B$ have the same probability of winning.
2. $A$ and $B$ are each twice as likely to win as $C$.

Thus: $P(A) = P(B) = 2p$ $P(C) = p$

The total probability of all three students winning must equal 1: $P(A) + P(B) + P(C) = 1$

Substitute the values: $2p + 2p + p = 1$ $5p = 1$ $p = \frac{1}{5}$

So: $P(A) = P(B) = 2p = \frac{2}{5}, \quad P(C) = p = \frac{1}{5}$

Step 2: Probability that $B$ or $C$ wins

The probability of $B$ or $C$ winning is: $P(B \cup C) = P(B) + P(C)$ $P(B \cup C) = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$

Final Answer:

The probability that $B$ or $C$ wins is: $\boxed{\frac{3}{5}}$

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Do you want a detailed explanation or additional examples?
Here are 5 related questions for practice:

1. If $A$ is three times as likely to win as $C$, and $B$ is twice as likely as $C$, what is the probability that $A$ or $C$ wins?
2. In a race with four participants $A, B, C, D$, where the probabilities are proportional to 1:2:3:4, calculate the probability of $B$ or $D$ winning.
3. What is the probability that neither $B$ nor $C$ wins in the original problem?
4. How does the total probability distribution change if $A$'s probability is halved while keeping $B$ and $C$'s proportions the same?
5. What is the probability of $A$ winning in the original problem setup?

Tip: Always check that the sum of probabilities equals 1 to ensure consistency.