
If the probability of the occurrence of a multiple of 3 on one die and a multiple of 2 on the other die if both are thrown together is $\dfrac{{11}}{k}$, then find k.
Answer
162.3k+ views
Hint: First. Find the sample space and count the total number of possible outcomes. Then find the favorable outcomes and the number of favorable outcomes. Using this information, find the probability.
Formula Used: \[{\text{Probability of an event = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Complete step by step Solution:
Sample space - \[ \begin{array}{*{20}{c}}
{\left( {1,1} \right),}&{\left( {1,2} \right),} \\
{\left( {2,1} \right),}&{\left( {2,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {1,3} \right),}&{\left( {1,4} \right),} \\
{\left( {2,3} \right),}&{\left( {2,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {1,5} \right),}&{\left( {1,6} \right),} \\
{\left( {2,5} \right),}&{\left( {2,6} \right),}
\end{array} \\
\begin{array}{*{20}{c}}
{\left( {3,1} \right),}&{\left( {3,2} \right),} \\
{\left( {4,1} \right),}&{\left( {4,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {3,3} \right),}&{\left( {3,4} \right),} \\
{\left( {4,3} \right),}&{\left( {4,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {3,5} \right),}&{\left( {3,6} \right),} \\
{\left( {4,5} \right),}&{\left( {4,6} \right),}
\end{array} \\
\begin{array}{*{20}{c}}
{\left( {5,1} \right),}&{\left( {5,2} \right),} \\
{\left( {6,1} \right),}&{\left( {6,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {5,3} \right),}&{\left( {5,4} \right),} \\
{\left( {6,3} \right),}&{\left( {6,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {6,5} \right),}&{\left( {5,6} \right),} \\
{\left( {6,6} \right),}&{\left( {6,6} \right)}
\end{array} \]
Favourable outcomes - $\left\{ {\left( {2,3} \right),\left( {2,6} \right),\left( {4,3} \right),\left( {4,6} \right),\left( {6,3} \right),\left( {6,6} \right),\left( {3,2} \right),\left( {6,2} \right),\left( {3,4} \right),\left( {6,4} \right),\left( {3,6} \right)} \right\}$
Number of favorable outcomes = 11
Probability of occurrence of a multiple of 3 on one die and a multiple of 2 on the other if both are thrown together is, therefore, $\dfrac{{11}}{{36}}$ because there are 11 favorable outcomes and 36 outcomes in total.
Therefore, the value of $k$ is 36
Note: Alternate way of solving – Probability of occurrence of a multiple of 3 on the first die is $\dfrac{1}{2}$ and probability of occurrence of a multiple of 2 on the second die is $\dfrac{1}{3}$. Therefore, the probability of getting a multiple of 3 on the first die and a multiple of 2 on the second die simultaneously is $\dfrac{1}{6}$. Similarly, the probability of getting a multiple of 2 on the first die and a multiple of 3 on the second die simultaneously is also $\dfrac{1}{6}$. Adding both of these we get $\dfrac{1}{3}$. However, the (6,6) occurs twice. Therefore, we need to subtract $\dfrac{1}{{36}}$ from $\dfrac{1}{3}$. Therefore, the answer is $\dfrac{{11}}{{36}}$.
Formula Used: \[{\text{Probability of an event = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Complete step by step Solution:
Sample space - \[ \begin{array}{*{20}{c}}
{\left( {1,1} \right),}&{\left( {1,2} \right),} \\
{\left( {2,1} \right),}&{\left( {2,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {1,3} \right),}&{\left( {1,4} \right),} \\
{\left( {2,3} \right),}&{\left( {2,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {1,5} \right),}&{\left( {1,6} \right),} \\
{\left( {2,5} \right),}&{\left( {2,6} \right),}
\end{array} \\
\begin{array}{*{20}{c}}
{\left( {3,1} \right),}&{\left( {3,2} \right),} \\
{\left( {4,1} \right),}&{\left( {4,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {3,3} \right),}&{\left( {3,4} \right),} \\
{\left( {4,3} \right),}&{\left( {4,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {3,5} \right),}&{\left( {3,6} \right),} \\
{\left( {4,5} \right),}&{\left( {4,6} \right),}
\end{array} \\
\begin{array}{*{20}{c}}
{\left( {5,1} \right),}&{\left( {5,2} \right),} \\
{\left( {6,1} \right),}&{\left( {6,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {5,3} \right),}&{\left( {5,4} \right),} \\
{\left( {6,3} \right),}&{\left( {6,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\left( {6,5} \right),}&{\left( {5,6} \right),} \\
{\left( {6,6} \right),}&{\left( {6,6} \right)}
\end{array} \]
Favourable outcomes - $\left\{ {\left( {2,3} \right),\left( {2,6} \right),\left( {4,3} \right),\left( {4,6} \right),\left( {6,3} \right),\left( {6,6} \right),\left( {3,2} \right),\left( {6,2} \right),\left( {3,4} \right),\left( {6,4} \right),\left( {3,6} \right)} \right\}$
Number of favorable outcomes = 11
Probability of occurrence of a multiple of 3 on one die and a multiple of 2 on the other if both are thrown together is, therefore, $\dfrac{{11}}{{36}}$ because there are 11 favorable outcomes and 36 outcomes in total.
Therefore, the value of $k$ is 36
Note: Alternate way of solving – Probability of occurrence of a multiple of 3 on the first die is $\dfrac{1}{2}$ and probability of occurrence of a multiple of 2 on the second die is $\dfrac{1}{3}$. Therefore, the probability of getting a multiple of 3 on the first die and a multiple of 2 on the second die simultaneously is $\dfrac{1}{6}$. Similarly, the probability of getting a multiple of 2 on the first die and a multiple of 3 on the second die simultaneously is also $\dfrac{1}{6}$. Adding both of these we get $\dfrac{1}{3}$. However, the (6,6) occurs twice. Therefore, we need to subtract $\dfrac{1}{{36}}$ from $\dfrac{1}{3}$. Therefore, the answer is $\dfrac{{11}}{{36}}$.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

IIIT JEE Main Cutoff 2024

IIT Full Form

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

JEE Main Cut-Off for NIT Kurukshetra: All Important Details

JEE Main Cut-Off for VNIT Nagpur 2025: Check All Rounds Cutoff Ranks

Other Pages
NEET 2025: All Major Changes in Application Process, Pattern and More

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025: Important Information and Key Updates

1 Billion in Rupees - Conversion, Solved Examples and FAQs

NEET 2025 Syllabus PDF by NTA (Released)

Important Days In June: What Do You Need To Know
