Suppose we have two dice one fair and one tha brings $6$ with quintuple probability.Find the probability to throw randomly one die and show $6$ Suppose we have two dice one fair and one that brings $6$ with quintuple probability than the other numbers.We get a die randomly and we throw it.What is the probability to have $6$?In the same problem if we know that we have $6$ what is the probability that we have throwed the second die?
Any ideas for these  parts especially the second one?
 A: The probability of rolling 6 on the fair die is obviously $\frac{1}{6}$.
Let $x$ denote the probability of rolling 6 on the non-fair die:


*

*Then $\frac{1}{5}x$ is the probability of rolling each one of the other $5$ values

*Therefore $x+5\cdot\frac{1}{5}x=1$, therefore $2x=1$, therefore $x=\frac{1}{2}$


So the probability of rolling 6 on the non-fair die is $\frac{1}{2}$.


What is the probability of rolling 6?

Split it into disjoint events, and then add up their probabilities:


*

*The probability of choosing the fair die and then rolling 6 on that die is $\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}$

*The probability of choosing the non-fair die and then rolling 6 on that die is $\frac12\cdot\frac{1}{2}=\frac{1}{4}$


So the probability of rolling 6 is $\frac{1}{12}+\frac{1}{4}=\frac{1}{3}$.


If we know that we rolled 6, what is the probability that it was on the non-fair die?

Use Bayes formula for conditional probability:


*

*Let $A$ denote the event of rolling the non-fair die

*Let $B$ denote the event of rolling 6


So $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{12}+\frac{1}{4}}=\frac{3}{4}$.
A: HINT: put a name at each random variable. By example $D\in\{d1,d2\}$ is the random variable of picking one dice or the other, and $T\in\{1,2,...,6\}$ is the random variable to get some value on the throw.
And we know that 
$$\Pr[T=6|D=d1]=1/6\quad\text{ and }\quad\Pr[T=6|D=d2]=1/2$$
Now we want to evaluate
1) $\Pr[T=6]$, and
2) $\Pr[D=d2|T=6]$
