Three students A,B and C are running a race. | vedclass

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(B) Let the winning probability of $C$ be $P(C) = p$. Since $A$ and $B$ are twice as likely to win as $C$,we have $P(A) = 2p$ and $P(B) = 2p$. Since t

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$\frac{3}{5}$

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(B) Let the winning probability of $C$ be $P(C) = p$. Since $A$ and $B$ are twice as likely to win as $C$,we have $P(A) = 2p$ and $P(B) = 2p$. Since the sum of probabilities of all possible outcomes is $1$,we have $P(A) + P(B) + P(C) = 1$. Substituting the values,we get $2p + 2p + p = 1$,which implies $5p = 1$,so $p = \frac{1}{5}$. The probability that $B$ or $C$ wins is $P(B \cup C) = P(B) + P(C)$ (since the events are mutually exclusive). $P(B \cup C) = 2p + p = 3p = 3 	imes \frac{1}{5} = \frac{3}{5}$.

Three students $A$,$B$ and $C$ are running a race. $A$ and $B$ have the same probability of winning and each is twice as likely to win as $C$. Then,the probability that $B$ or $C$ wins is equal to (assuming there are no ties)

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