Question

In a Derby race, the horse Cusac's chance of winning is \[\dfrac{1}{6}\], that of Lydo's winning is \[\dfrac{1}{12}\] and of horse Delco's is \[\dfrac{1}{8}\]. If the race starts with \[18\] horses and only one horse wins, what is the probability that one of these \[3\] horses will win?

Answer 1

Hint: In this question it is given that only one horse can win the race so it makes the three events mutually exclusive which means if one event happens at the end of the experiment then the other two events won't happen. We are going to use the rule of addition for three events and the formula is, \[P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( C\cap A \right)+P\left( A\cap B\cap C \right)\], where \[P\left( A \right)\] is the probability of winning of the horse Cusac (event A), \[P\left( B \right)\] is the probability of winning of the horse Lydo (event B) and \[P\left( C \right)\] is the probability of winning of horse Delco (event C). All the terms with intersection symbols will become zero because it is given that only one event can happen as they all are mutually exclusive.

Complete step-by-step answer:
In this question we have to find the probability of winning one of the horses and it is given that only one horse wins the race at the end.
Let \[P\left( A \right)\] be the probability of winning of the horse Cusac (event A), \[P\left( B \right)\] is the probability of winning of the horse Lydo (event B) and \[P\left( C \right)\] is the probability of winning of horse Delco (event C).
It is given that these three events are mutually exclusive which means that only one event will occur at the end of the experiment and the other two will not occur.
Mathematically it means that the terms with intersection symbol will be equal to zero because intersection symbol \[\cap \] represents 'multiplication' as well as 'and' which means taking all the events together or working on the common outputs of all events in the experiment, \[A\cap B\] is \[A\] and \[B\] whereas union symbol \[\cup \] represents 'addition' as well as 'or' which means taking either one of all the events, \[A\cup B\] is either \[A\] or \[B\].
Mathematically mutually exclusive events mean,
\[\begin{align}
  & P\left( A\cap B \right)=0 \\
 & P\left( B\cap C \right)=0 \\
 & P\left( C\cap A \right)=0 \\
 & P\left( A\cap B\cap C \right)=0 \\
\end{align}\]
Probability of occurring of multiple or all the events together at once is 0.
In the question is it given all the probabilities of the events which are,
\[P\left( A \right)=\dfrac{1}{6}\], Probability of winning of the horse Cusac (event A),
\[P\left( B \right)=\dfrac{1}{12}\], Probability of winning of the horse Lydo (event B),
\[P\left( C \right)=\dfrac{1}{8}\], Probability of winning a horse Delco (event C).
Now putting the known values in the formula for rule of addition which is, \[P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( C\cap A \right)+P\left( A\cap B\cap C \right)\], we get,
 \[P\left( A\cup B\cup C \right)=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{8}-0-0-0+P\left( A\cap B\cap C \right)\]
Now even the intersection term of all the three events will become zero because not all the three horses can win the race in one experiment. We will get,
\[P\left( A\cup B\cup C \right)=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{8}\]
\[P\left( A\cup B\cup C \right)=\dfrac{4+2+3}{24}\]
\[P\left( A\cup B\cup C \right)=\dfrac{9}{24}=\dfrac{3}{8}\]
So the final answer is \[P\left( A\cup B\cup C \right)=\dfrac{3}{8}\] which is the probability that one of the three horses will win the race.

Note: First of all we need to remember the formula of rule of addition for two events as well as three events by using the Venn diagrams given in the set theory. Remember all the definitions of different types of events like compound events, mutually exclusive events, exhaustive events and many more. In these kinds of questions there is no need to construct a sample space as one would not be able to do so and it is also not needed as the probabilities are given for each event and we have to calculate the probability of occurrence of any one event. But remember whenever possible always construct the sample space before attempting the questions of probability.