What is the probability of these questions?

Question

What is the probability of throwing $2$ dice, and the first one is a $6$? (Given that one of them is a 6.)

My solution says that it's $\frac{6}{11}$, but I have no clue why.

The other one is:

What is the probability of both is a $6$? ($\frac{1}{11}$)

Answer 1

Given the answers, the question that was likely intended to be asked is the following:

Having thrown two fair standard six-sided dice one at a time, and given that at least one of the dice shows a six, what is the probability that

• The first die thrown shows a six
• Both dice thrown show a six

Let $X$ be the random variable denoting the result of the first die. Let $Y$ be the random variable denoting the result of the second die.

Reworded then, the problem asks us to find $Pr(X=6\mid (X=6)~\text{or}~(Y=6))$ as well as $Pr((X=6)~\text{and}~(Y=6)\mid (X=6)~\text{or}~(Y=6))$

Letting $A$ be the event that $X=6$ and $B$ be the event that $Y=6$, reworded yet again, this is asking us to find $Pr(A\mid A\cup B)$ and $Pr(A\cap B\mid A\cup B)$

Conditional probabilities: $$Pr(E\mid F) = \frac{Pr(E\cap F)}{Pr(F)}$$ In the special case of working in an equiprobable sample space, this simplifies to $$Pr(E\mid F) = \frac{|E\cap F|}{|F|}$$

There are eleven possibilities for having at least one of the dice to show a six. This can be seen as $|A\cup B| = |A|+|B|-|A\cap B| = 6+6-1=11$

Of those eleven possibilities, each are equally likely. They are specifically $(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),\dots,(6,6)$

Six of said eleven possibilities will have the first die showing a six. $(6,1),(6,2),\dots,(6,6)$. That is, $|A\cap (A\cup B)|=|A|=6$

We get then, $Pr(A\mid A\cup B) = \frac{|A|}{|A\cup B|}=\frac{6}{11}$

Similarly, only one of the eleven outcomes is with both dice showing a six, so $Pr(A\cap B\mid A\cup B) = \frac{1}{11}$

Note: these are very different from the question of finding $Pr(A)$. In general, $Pr(A)\neq Pr(A\mid B)$ except in very special circumstances (when $A$ and $B$ are independent events).

Answer 2

If you know that at least one of them is a 6, you get a total of $11$ possibilities $(1,6),(2,6),...(6,6)$ and $(6,1),(6,2)...(6,5)$ (you canot repeat the 6,6). You just have to count. The possibility of having a 6 on the first dice, knowing you got at least one is $6/11$

And the probability of two sixes knowing that there is at least one is $1/11$