An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four values obtained the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 is
A. $\dfrac{16}{81}$
B. $\dfrac{1}{81}$
C. $\dfrac{80}{81}$
D. $\dfrac{65}{81}$
Answer
531k+ views
Hint:We first explain the concept of empirical probability and how the events are considered. We take the given events and find the number of outcomes. Using the probability theorem of $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$, we get the empirical probability of the shooting event.
Complete step by step answer:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome. We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options. We take the conditional event A as getting four rolls with the minimum face value being not less than 2 and the maximum face value not being greater than 5 and the universal event U as rolling four times and numbers will be denoted as $n\left( A \right)$ and $n\left( U \right)$.
We take the empirical probability of the given problem as,
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$
For $n\left( U \right)$, the four rolls can have any number in between 1 to 6. So, every roll has 6 options. So, the number of outcomes is $n\left( U \right)={{6}^{4}}$. For $n\left( A \right)$, the four rolls can have any number in between 2 to 5. So, every roll has 4 options. So, the number of outcomes is $n\left( A \right)={{4}^{4}}$.
The empirical probability of the shooting event is,
\[P\left( A \right)=\dfrac{{{4}^{4}}}{{{6}^{4}}} \\
\Rightarrow P\left( A \right) =\dfrac{{{2}^{4}}}{{{3}^{4}}} \\
\therefore P\left( A \right) =\dfrac{16}{81}\]
Hence, the correct option is A.
Note:We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact.
Complete step by step answer:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome. We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options. We take the conditional event A as getting four rolls with the minimum face value being not less than 2 and the maximum face value not being greater than 5 and the universal event U as rolling four times and numbers will be denoted as $n\left( A \right)$ and $n\left( U \right)$.
We take the empirical probability of the given problem as,
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$
For $n\left( U \right)$, the four rolls can have any number in between 1 to 6. So, every roll has 6 options. So, the number of outcomes is $n\left( U \right)={{6}^{4}}$. For $n\left( A \right)$, the four rolls can have any number in between 2 to 5. So, every roll has 4 options. So, the number of outcomes is $n\left( A \right)={{4}^{4}}$.
The empirical probability of the shooting event is,
\[P\left( A \right)=\dfrac{{{4}^{4}}}{{{6}^{4}}} \\
\Rightarrow P\left( A \right) =\dfrac{{{2}^{4}}}{{{3}^{4}}} \\
\therefore P\left( A \right) =\dfrac{16}{81}\]
Hence, the correct option is A.
Note:We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact.
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