Question

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?

Answer 1

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 \[1.\]