Answer
Verified
384.6k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE