# 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

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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 \[1.\]

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 \[1.\]

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