Throwing 8 sided dice 3 times with the sum of the throws being 6, whats the probability that the first throw was 3?

Question

We already threw the 8 sided dice 3 times. The sum of these 3 throws was 6. Whats the probability that in our first throw we got 3?

I am a little lost on this one, but I would probably do it something like this:

3 _ _ -> 3 2 1 | 3 1 2 which is the 2 ways that we are trying to find the probability of happening. All possible ways to get sum of 6 would be: _ _ _ -> 2 2 2 -> 2 3 1 | 2 1 3 | 1 2 3 | 1 3 2 | 3 2 1 | 3 1 2 | 4 1 1 | 1 4 1 | 1 1 4. So is the probability 2/10?

Answer 1

This is conditional probability. If $A$ is the event that the first throw was $3$ and $B$ is the event that the sum is $6$, then you are looking for $P(A|B)$.

Using Bayes' Theorem:

$$ P(A|B) = \frac{P(A)P(B|A)}{P(B)}$$

$$P(A) = \frac{1}{8}$$

$$P(B|A) = P(sum\quad of\quad two\quad throws\quad is\quad 3) = \frac{2}{64}$$

$$P(B) = \frac{10}{512}$$

Thus

$$P(A|B) = \frac{2}{10} = \frac{1}{5}$$

Answer 2

You've missed the combinations 4 1 1, 1 4 1, 1 1 4 There'll be a total of 10 combinations with seven that you've mentioned Thus the probability will be 2/10

Answer 3

The first roll is $3$ in $2$ out of $10$ cases (of the sum of $3$ rolls being $6$):

• $1,1,4$
• $1,2,3$
• $1,3,2$
• $1,4,1$
• $2,1,3$
• $2,2,2$
• $2,3,1$
• $\color\red{3,1,2}$
• $\color\red{3,2,1}$
• $4,1,1$

So the probability is $\dfrac{2}{10}$