Question

In a swimming competition, only 3 students A, B, and C are taking part. The probability of A’s winning or B’s winning is three times the probability of C’s winning. The probability of the event either B or C to win is
(a) \[\dfrac{1}{7}\]
(b) \[\dfrac{2}{7}\]
(c) \[\dfrac{3}{7}\]
(d) \[\dfrac{4}{7}\]

Answer 1

Hint: We solve this problem by using the probability condition.
The probability condition is that the sum of all probabilities of individual events is equal to 1. That is if \[P\left( A \right),P\left( B \right),P\left( C \right)\] are the probabilities of winning of A, B and C then
\[P\left( A \right)+P\left( B \right)+P\left( C \right)=1\]
Now we use the condition that the winning events of A, B, and C are independent events.
We use the condition that the combined probability of independent events is obtained by adding the probabilities of independent events. That is if E is the probability of winning B or C then
\[P\left( E \right)=P\left( B \right)+P\left( C \right)\]

Complete step by step answer:
We are given that the probability of A’s winning or B’s winning is three times the probability of C’s winning.
Let us assume that \[P\left( A \right),P\left( B \right),P\left( C \right)\] are the probabilities of winning of A, B and C
Now, by converting the given statement into mathematical equation we get
\[\Rightarrow P\left( A \right)=P\left( B \right)=3P\left( C \right)\]
We know that the probability condition is that the sum of all probabilities of individual events is equal to 1. That is if \[P\left( A \right),P\left( B \right),P\left( C \right)\] are the probabilities of winning of A, B and C then
\[P\left( A \right)+P\left( B \right)+P\left( C \right)=1\]
Now, by substituting the required values in above equation that is substituting \[P\left( A \right),P\left( B \right)\] in terms of \[P\left( C \right)\] in above equation we get
\[\begin{align}
  & \Rightarrow 3P\left( C \right)+3P\left( C \right)+P\left( C \right)=1 \\
 & \Rightarrow 7P\left( C \right)=1 \\
 & \Rightarrow P\left( C \right)=\dfrac{1}{7} \\
\end{align}\]
Now, by substituting the value of \[P\left( C \right)\] in \[P\left( A \right),P\left( B \right)\] we get
\[\Rightarrow P\left( A \right)=P\left( B \right)=\dfrac{3}{7}\]
Here, we can see that the events of winning of A, B and C are independent events because winning of one person doesn’t affect the winning of other persons.
Let us assume that E be the event of winning of B or C
We know that the combined probability of independent events is obtained by adding the probabilities of independent events.
By using this condition to event E we get
\[P\left( E \right)=P\left( B \right)+P\left( C \right)\]
By substituting the required probabilities in above equation we get
\[\begin{align}
  & \Rightarrow P\left( E \right)=\dfrac{3}{7}+\dfrac{1}{7} \\
 & \Rightarrow P\left( E \right)=\dfrac{4}{7} \\
\end{align}\]
Therefore we can conclude that the probability of winning of B or C is \[\dfrac{4}{7}\]
So, option (d) is correct answer.


Note:
Students may do mistakes in taking the probability of combined events.
Here we are asked to find the probability of a combined event that is winning of B or C
We know that in a probability the or indicates the union and we have the probability of the union of events as
\[P\left( B\cup C \right)=P\left( B \right)+P\left( C \right)-P\left( B\cap C \right)\]
But here the events of winning A, B, and C are independent events so that there will be no intersection. So, we take the formula as
\[P\left( B\cup C \right)=P\left( B \right)+P\left( C \right)\]
But students do not consider the independent events and take the formula as
\[P\left( B\cup C \right)=P\left( B \right)+P\left( C \right)-P\left( B\cap C \right)\]
Here, we cannot find the probability because we have no condition to find the intersection of events.