
If a fair 6-sided dice is rolled three times, what is the probability that exactly one three is rolled.
Answer
137.4k+ views
Hint: First we calculate the total number of possible outcomes. Then, we consider the case that exactly one three is rolled and find the number of favorable outcomes. Then, we calculate the probability by using the general formula of probability which is given by
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$.
Complete step-by-step solution:
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
We have given that a fair 6-sided dice is rolled three times.
We have to find the probability that exactly one three is rolled.
Now, the total number of ways in which 6-sided dice is rolled three times will be $6\times 6\times 6=216$
So, we have total number of possible outcomes $n\left( S \right)=216$
Now, there is three possible cases that exactly one three is rolled which are as following:
$\begin{align}
& 3,N,N \\
& N,3,N \\
& N,N,3 \\
\end{align}$
It means three on the first roll and any other numbers on two rolls. So, the possible cases will be $1\times 5\times 5=25$
Now, three on the second roll and any other number on two rolls, so the possible cases will be $5\times 1\times 5=25$
Now, three on the third roll and any other number on two rolls, so the possible cases will be
$5\times 5\times 1=25$
Now, above three cases are separate and independent to each other, so we have to add the probabilities together, we get
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
\[\begin{align}
& P\left( A \right)=\dfrac{25}{216}+\dfrac{25}{216}+\dfrac{25}{216} \\
& P(A)=\dfrac{75}{216} \\
& P(A)=\dfrac{25}{72} \\
\end{align}\]
So, the probability of getting exactly one three is $\dfrac{25}{72}$.
Note: When we have to calculate the probability of a series of events, where events are separate and independent to each other, we must add the separate probabilities of events together to get the combined probability.
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$.
Complete step-by-step solution:
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
We have given that a fair 6-sided dice is rolled three times.
We have to find the probability that exactly one three is rolled.
Now, the total number of ways in which 6-sided dice is rolled three times will be $6\times 6\times 6=216$
So, we have total number of possible outcomes $n\left( S \right)=216$
Now, there is three possible cases that exactly one three is rolled which are as following:
$\begin{align}
& 3,N,N \\
& N,3,N \\
& N,N,3 \\
\end{align}$
It means three on the first roll and any other numbers on two rolls. So, the possible cases will be $1\times 5\times 5=25$
Now, three on the second roll and any other number on two rolls, so the possible cases will be $5\times 1\times 5=25$
Now, three on the third roll and any other number on two rolls, so the possible cases will be
$5\times 5\times 1=25$
Now, above three cases are separate and independent to each other, so we have to add the probabilities together, we get
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
\[\begin{align}
& P\left( A \right)=\dfrac{25}{216}+\dfrac{25}{216}+\dfrac{25}{216} \\
& P(A)=\dfrac{75}{216} \\
& P(A)=\dfrac{25}{72} \\
\end{align}\]
So, the probability of getting exactly one three is $\dfrac{25}{72}$.
Note: When we have to calculate the probability of a series of events, where events are separate and independent to each other, we must add the separate probabilities of events together to get the combined probability.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

How to find Oxidation Number - Important Concepts for JEE

Half-Life of Order Reactions - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Collision - Important Concepts and Tips for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 13 Statistics

NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines
