Question

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 1

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 $\dfrac{{11}}{{36}}$.