
If a fair 6-sided dice is rolled three times, what is the probability that exactly one three is rolled.
Answer
232.8k+ 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.
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