I am reading the following question:
What is the probability of throwing precisely one six when 6 dice are rolled?
So when we have one dice the probability of having one six is $\frac{1}6$
When we have six dice we can have 6 cases that the six can come i.e. first dice or second or third etc.
The total possible outcomes of throwing 6 dice is: $6\cdot6\cdot6\cdot6\cdot6\cdot6=6^6$
So the probability is: $\frac{6}{6^6}=\frac{1}{6^5}$
But the solution says:
The probability of throwing precicely $1$ six out of $6$ dices is:
$\binom{6}{1}\cdot\frac{1}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6$
Could someone help me understand why is $\binom{6}{1}$ being used here? That equals to $6$ as I also reasoned but why would we use choice formula here? What does it mean?
Also I assume the $\frac{5}6$ is the probability of not throwing a $6$ but why does it have to be multiplied $6$ times.
If I apply this formula to the case of throwing $1$ six when we roll $2$ dice I get:
$\binom{2}{1}\cdot\frac{1}6\cdot\frac{5}6\cdot\frac{5}6 = 2\cdot\frac{1}{6}\cdot\frac{5^2}{6^2}=\frac{25}{108}$
But if I actually count the cases we have:
| Dice 1 | Dice 2 |
|---|---|
| 1 | 6 |
| 2 | 6 |
| 3 | 6 |
| 4 | 6 |
| 5 | 6 |
| 6 | 1 |
| 6 | 2 |
| 6 | 3 |
| 6 | 4 |
| 6 | 5 |
this is $\frac{10}{36}$.
So what am I doing wrong here?
Expanding what @N.F.Taussig said in the comments of the question.
Instead of thinking about $6$ dice think about one single dice rolled six times. In six rolls you want to have one exaclty $6$ and all the other results must be different.
You can get a $6$ on the first throw with probability $\frac{1}{6}$ and then the probability that in the next $5$ rolls you do not get a $6$ is $\left(\frac{5}{6}\right)^5$
On the opposite hand, at the first throw you can get a number different from $6$ with probability $5/6$ and get a $6$ at the second throw. Then all the other 4 throws must end up with a number different from a $6$. This gives you again $\frac{1}{6}\left(\frac{5}{6}\right)^5$
Go on and count the probability you can get the $6$ at the $i$-th throw for $i = 3,4,5,6$.
This lead you to the result that the probability of getting exactly one $6$ in $6$ different rolls is $$\binom{6}{1}\cdot\frac{1}{6}\left(\frac{5}{6}\right)^5$$
If the book gives $\left(\frac{5}{6}\right)^6$ then it is an error.