Finding the probability of rolling sixes in both die Hi guys I am writing my P exam for the second time and I remembered two question that confused me when writing the exam. I asked my prof. but it confused him as well. For simplicity I will ask one question here and post the other one after.
So if you could please help me I would really be greatful.
Question:
You have four die in a urn. 2 regular die, 1 die with faces $(2,2,6,6,6,5)$, and 1 die with all 6.
You throw 2 die. What is the probability of getting 6 on both.
My attempt:
I labelled the dies A,B,C,D respectively as in the question. Then i thought to my self there are total of 9 ways for these to occur. but without repetition like AB is the same as BA. so I got the choices $AB, AC, AD, BC, BD, CD$ and the probability of these respectively is $(\frac{1}{36}, \frac{1}{12}, \frac{1}{6}, \frac{1}{12}, \frac{1}{6}, \frac{1}{2})$
I dont know what to do after. do I add them all.
Really confused please help out.
Thank you
 A: Why don't you think of it in terms of conditional probabilities such as $P(\text{two sixes}|\text{dice are}\; AB)$? 
So compute 
$$P(2\; \text{sixes}|AB)P(AB) + P(2\; \text{sixes}|AC)P(AC) + \ldots + P(2\; \text{sixes}|CD)P(CD)$$
where $P(AB) = \ldots = P(CD) = 1/6$. The conditional probability for the case $AD$ (for example) is $1/6 \cdot 1 = 1/6$. And so on. It's pretty easy from here. 
A: You are choosing two dice from the four in the urn. There are $_4 C_2=6$ ways to do this, all of which are equally likely, so the probability of choosing any particular pair of dice is $\frac{1}{6}$.
Then for each case, find the conditional probability of rolling two sixes. For example, given that you choose the two ordinary dice, the conditional probability that you roll two sixes is 
$\frac{1}{36}$. So the list of fractions you have there is a list of conditional probabilities of getting two sixes, given the particular choice of the two dice. To get the unconditional probability you want, simply multiply each by $\frac{1}{6}$ and sum. Or, what's the same, sum them and divide by 6.
Luckily it turns out that if you forget and just sum them, you get $\frac{37}{36}>1$, setting off alarm bells. Just divide by 6. The answer is $\frac{37}{216} \approx0.17$.
In symbols, the rule we're using here is that if $A_1$, $A_2$, …, $A_n$ are mutually exclusive events  that exhaust the sample space, then
$$P(B) = P(B \mid A_1) P(A_1) + P(B \mid A_2)P(A_2) + \cdots + P(B \mid A_n)P(A_n)
$$
for any event $B$. This is called the law of total probability.
Edit: This, of course, is if the two dice are chosen without replacement. In other words, if you first choose two dice and then roll them both. If you choose one die, roll it, put it back in the urn, and choose a die from the urn again, the answer will be different. This is because in the second case, the two sixes can come from the same die, while in the first case that's impossible. I read "you throw two dice" as meaning you roll two dice at the same time.
