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(D) Let the probability that $C$ wins be $P(C) = p$.According to the problem, the probability that $B$ wins is $P(B) = 2p$.The probability that $A$ wi

Vedclass · Accepted Answer

$\frac{3}{7}$

Vedclass · Answer

(D) Let the probability that $C$ wins be $P(C) = p$.According to the problem, the probability that $B$ wins is $P(B) = 2p$.The probability that $A$ wins is $P(A) = 2 	imes P(B) = 2(2p) = 4p$.Since one of them must win, the sum of their probabilities is $1$:$P(A) + P(B) + P(C) = 1$$4p + 2p + p = 1$$7p = 1 \Rightarrow p = \frac{1}{7}$.Therefore, the probability that $A$ wins is $P(A) = 4p = \frac{4}{7}$.The probability that $A$ loses is $P(\bar{A}) = 1 - P(A) = 1 - \frac{4}{7} = \frac{3}{7}$.

In a competition, $A, B$, and $C$ are participating. The probability that $A$ wins is twice that of $B$, and the probability that $B$ wins is twice that of $C$. What is the probability that $A$ loses?

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