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# Three horses $A,B$ and $C$ are in a race. $A$ is twice as likely to win as $B$ and $B$ is twice as likely to win as $C.$ What are their probabilities of winning? Verified
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Hint: Sum of all probabilities will be $1.$
Let the probabilities of winning the race by $A,B$ and $C$ be $P(A),P(B)$and $P(C)$respectively.
Then, total probability will be $1.$
$\Rightarrow P(A) + P(B) + P(C) = 1{\text{ }}.....(i)$
Now, according to the question, $A$ is twice as likely to win as $B$:
$\Rightarrow P(A) = 2P(B){\text{ }}.....(ii)$
And $B$ is twice as likely to win as $C$:
$\Rightarrow P(B) = 2P(C) \\ \Rightarrow P(C) = \frac{{P(B)}}{2}{\text{ }}.....(iii) \\$
Putting the values of $P(A)$ and $P(C)$from equations $(ii)$and $(iii)$ to $(i)$.
We’ll get:
$\Rightarrow 2P(B) + P(B) + \frac{{P(B)}}{2} = 1, \\ \Rightarrow \frac{7}{2}P(B) = 1, \\ \Rightarrow P(B) = \frac{2}{7}. \\$
Putting the value of $P(B)$ in equation $(ii)$and $(iii)$,
$\Rightarrow P(A) = 2P(B) = 2 \times \frac{2}{7}, \\ \Rightarrow P(A) = \frac{4}{7}. \\$
Similarly,
$\Rightarrow P(C) = \frac{{P(B)}}{2} = \frac{1}{2} \times \frac{2}{7}, \\ \Rightarrow P(C) = \frac{1}{7}. \\$
Thus, the required probabilities of winning the race by $A,B$ and $C$ are $\frac{4}{7},\frac{2}{7}$ and $\frac{1}{7}$respectively.
Note: Sum of probabilities of mutually exclusive events is always $1.$So in case if there were more than three horses, the sum of respective probabilities of winning of each horse would also have been $1.$