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In a musical chair game, the person playing the music has been advised to stop playing the music at any time within $2$ minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?

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Last updated date: 24th Jun 2024
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Answer
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Hint: The probability of an event is equal to the ratio of the number of the favorable outcomes to the total number of possible outcomes. In this case, the event is the stopping of music within the first half-minute. Also, it is given that the music can be stopped at any time within the first two minutes. So the possible outcomes are all the numbers lying between $0$ and $2$, while the favorable outcomes are all the numbers lying between $0$ and $\dfrac{1}{2}$.

Complete step by step solution:
Let $t=0$ be the time when the music is started.
Let us call the stopping of music within the first half-minute as the event $E$.
Since the person has to stop the music within the first two minutes, the outcome can be any time between $t=0$ and $t=2$ minutes. This means that the total outcomes or the sample space consists of all the numbers between $0$ and $2$. Also, the event $E$ corresponds to the stopping of music at any time between $t=0$ and $t=\dfrac{1}{2}$ minutes. So the favorable outcomes consists of all the numbers between $0$ and $\dfrac{1}{2}$.
The distance of $2$ from $0$ is equal to \[2\] units, and the distance of $\dfrac{1}{2}$ from $0$ is equal to $\dfrac{1}{2}$ units.
Now all numbers between $0$ and $2$ are equally likely to occur. So we can say that the distance of \[2\] units can be associated to the total number of outcomes, and the distance of $\dfrac{1}{2}$ units can be associated to the event $E$. Now, we know that the probability of an event is equal to the ratio of the number of the favorable outcomes to the total number of possible outcomes. So the probability of the event $E$ is given by
$\begin{align}
  & \Rightarrow p\left( E \right)=\dfrac{\dfrac{1}{2}}{2} \\
 & \Rightarrow p\left( E \right)=\dfrac{1}{4} \\
\end{align}$
Hence, the required probability is equal to $\dfrac{1}{4}$.

Note: In the case of the above question, there were infinitely many outcomes in the sample space. So we cannot count them to substitute them into the formula of probability. So we associated a countable quantity, distance, by using the fact that all of the infinite outcomes were equally likely to occur. Whenever there is a case of infinite outcomes, probability is calculated in this way only.