
Two dice are thrown simultaneously. Find the probability of getting: a multiple of 2 on one dice and a multiple of 3 on other.
Answer
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Hint: First we have to find the probability of individual events. Then, we have to evaluate the sample space of the particular event and then find the favorable outcome from the sample set and finally evaluate the probability.
Complete step-by-step answer:
If a random experiment is performed, then each of its outcomes is known as an elementary event.
The set of all possible outcomes of a random experiment is called the sample space associated with it and it is generally denoted by ‘S’.
For an event of throwing of single die, sample space of die is given as:
\[s\text{ }=\text{ }\left\{ 1,\text{ }2,\text{ }3,\text{ }4,\text{ }5,\text{ }6 \right\}.\]
Similarly, for an event of two dice thrown simultaneously, sample space of two dice are given as:
$\begin{align}
& S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\
& \text{ }(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
& \text{ }(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\
& \text{ }(4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\
& \text{ }(5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\
& \text{ }(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \\
\end{align}$
The space of the multiple of 2 on one dice and a multiple of 3 on other dice is given as:
$E=\{(2,3),(2,6),(4,3),(4,6),(6,3),(6,6)\}$
So, now the probability of the given event(E) can be shown as:
\[P(E)=\dfrac{(Number\text{ of favorable outcomes)}}{(T\text{otal number of possible outcomes)}}\]
Therefore, the number of favorable outcomes in the sample space = 6.
Total number of possible outcomes of sample space = 36.
The probability of the given event is:
P (Getting multiple of 2 on one dice and multiple of 3 on other dice)$=\dfrac{6}{36}$
P (Getting multiple of 2 on one dice and multiple of 3 on other dice)$=\dfrac{1}{6}.$
$\therefore $The probability of getting multiple of 2 on one dice and multiple of 3 on other dice is $\dfrac{1}{6}.$
Note: Focus must be emphasized on finding the sample space and favorable outcomes particularly for probability related problems. After this the problem is reduced to mathematical calculations only.
Complete step-by-step answer:
If a random experiment is performed, then each of its outcomes is known as an elementary event.
The set of all possible outcomes of a random experiment is called the sample space associated with it and it is generally denoted by ‘S’.
For an event of throwing of single die, sample space of die is given as:
\[s\text{ }=\text{ }\left\{ 1,\text{ }2,\text{ }3,\text{ }4,\text{ }5,\text{ }6 \right\}.\]
Similarly, for an event of two dice thrown simultaneously, sample space of two dice are given as:
$\begin{align}
& S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\
& \text{ }(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
& \text{ }(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\
& \text{ }(4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\
& \text{ }(5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\
& \text{ }(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \\
\end{align}$
The space of the multiple of 2 on one dice and a multiple of 3 on other dice is given as:
$E=\{(2,3),(2,6),(4,3),(4,6),(6,3),(6,6)\}$
So, now the probability of the given event(E) can be shown as:
\[P(E)=\dfrac{(Number\text{ of favorable outcomes)}}{(T\text{otal number of possible outcomes)}}\]
Therefore, the number of favorable outcomes in the sample space = 6.
Total number of possible outcomes of sample space = 36.
The probability of the given event is:
P (Getting multiple of 2 on one dice and multiple of 3 on other dice)$=\dfrac{6}{36}$
P (Getting multiple of 2 on one dice and multiple of 3 on other dice)$=\dfrac{1}{6}.$
$\therefore $The probability of getting multiple of 2 on one dice and multiple of 3 on other dice is $\dfrac{1}{6}.$
Note: Focus must be emphasized on finding the sample space and favorable outcomes particularly for probability related problems. After this the problem is reduced to mathematical calculations only.
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