Three athletes A, B and C participate in a ra | vedclass

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Question

(C) Let $P(A), P(B),$ and $P(C)$ be the probabilities of winning for athletes $A, B,$ and $C$ respectively.Given that $P(A) = 2P(C)$ and $P(B) = 2P(C)

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$\frac{4}{5}$

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(C) Let $P(A), P(B),$ and $P(C)$ be the probabilities of winning for athletes $A, B,$ and $C$ respectively.Given that $P(A) = 2P(C)$ and $P(B) = 2P(C)$.Since the sum of probabilities of all mutually exclusive outcomes is $1$,we have:$P(A) + P(B) + P(C) = 1$Substituting the values,we get:$2P(C) + 2P(C) + P(C) = 1$$5P(C) = 1 \Rightarrow P(C) = \frac{1}{5}$Thus,$P(A) = \frac{2}{5}$ and $P(B) = \frac{2}{5}$.The probability that the race is won by $A$ or $B$ is $P(A \cup B) = P(A) + P(B)$.$P(A \cup B) = \frac{2}{5} + \frac{2}{5} = \frac{4}{5}$.

List $I$	List $II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

Three athletes $A, B$ and $C$ participate in a race competition. The probability of winning for $A$ and $B$ is twice the probability of winning for $C$. Then the probability that the race is won by $A$ or $B$ is:

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