Conditional probablity loaded die problem

[raptor1770]

This is part of the study guide for an exam and I'm not sure how to start.

1. The problem statement, all variables and given/known data

Consider three identical-looking dice. Two of the dice are ordinary fair dice (six equally-likely faces, numbered 1,2,...,6), but the third die is "loaded" (the face that ordinarily has a 1 has a 6 instead; that is, the loaded die has two 6's, on opposite faces). Fran and Ron each choose a die at random, and the remaining die is discarded.
Suppose that Fran and Ron roll their dice simultaneously

1. Find the probability that Fran rolls a 6.


4. Find the probability that Fran rolls a 6 if it is known that Ron rolled a 6.

Am I supposed to use Baye's theorem to solve this problem? Thank you for any help.

[verty]

You have not made an attempt to solve the problem.

[raptor1770]

Well I'm really at a loss as to where to start but I'll give you what I got so far.

The probability of rolling a six on a normal die is 1/6
on a fixed die it is 1/3

The probability that Fran picked the loaded die is 1/3; the regular die 2/3.

So the total probability of rolling a 6 is probability of 6 on loaded multiplied by chance to pick loaded plus probability of 6 on unloaded die multiplied by chance to pick unloaded.

or

(1/3*1/3) + (2/3*1/6) = .2222

Let me know if I am on the right track. Thank you.

EDIT: Still working on 4.

[dacruick]

looks good. Make sure to remember that there are different ways in which number 4 can happen. this depends on if ron has the loaded die or not.

[raptor1770]

For question 4 I think the solution is something like this.

Let L = Loaded
U = Unloaded
p = probability

The probability of rolling 6 given 6 is

= \frac{p(6|6) p(6)}{p(6|L) p(L) + p(6|U) p(U)}

This is derived using Baye's theorem. I think the numerator is incorrect because I dont know what the probability of 6 given a roll of 6 is.

[dacruick]

if you do the math out for me then I'll guide you. That really doesn't help me troubleshoot your method.

What I would suggest doing is finding the probability of rolling a 6 given that Ron rolled the loaded die, and then find the probability of rolling a 6 if Ron rolled a regular die.

Then there is another step but I think you can get it. Its actually very very similar to question 1.

[raptor1770]

OK based on your suggestion I calculated the probablity of rolling a 6 based on Ron getting the loaded die and the unloaded die. I think it's right please let me know.

If Ron got the loaded die:
(1/6 * 1) + (1/3 * 0) = 1/6

If Ron got the unloaded die:
(1/6 * 1/2) + (1/3 * 1/2) = 1/4

So from that information I can calculate the total probablity which is:
The probability of rolling a 6 if Ron got the loaded multiplied by the probability of Ron getting a loaded die, plus the probability of rolling a 6 if Ron got the unloaded multiplied by the probability of Ron getting the unloaded or:

(1/6)*(1/3) + (1/4)*(2/3) = .222

Thanks

[dacruick]

perfect.