Need help with Conditional Statistics

1. Oct 18, 2012

JP16

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

1) S'pose we flip 2 fair coins, and roll one fair 6 sided die. What is the probability that the number of heads equals the number showing on the die?

2) We roll 4 fair six sided dice. What is the conditional prob. that the first die shows 2, conditional on the event that at LEAST 3 dice show 2.

2. Relevant equations

P(A|B) = P(A and B)/ P(B)

3. The attempt at a solution

1) 0.25 / (1/3) = 3/4. Is this correct? I have my doubts. :|

EDIT: Now I think it is: (1/3) / 0.75 = 4/9

2) The first of the question asks, if "EXACTLY" 3 dice show 2, in which the answer is 3/4. But for at least 3 show 2, I don't get how to mathematically show it, using the above formula. Intuitively I understand that the prob. has to be greater than 3/4 in this case, just don't know how to show. Any help would be appreciated.

Last edited: Oct 18, 2012
2. Oct 18, 2012

Ray Vickson

For 1): don't just write down some numbers as a final answer; show how you obtained it. The same advice applies to 2).

RGV

3. Oct 18, 2012

JP16

Thank for the quick reply. For 1)

Denote by A, the # of heads.
by B the # on dice.

Then P(B|A) = P(B and A)/ P (A) - For P(A), for at least 1 heads the prob. is 3/4. For intersection, possible sample space is for dice to be either 1 or 2 (1/3). But i'm confused as to what is the intersection of A and B. I am thinking since A and B are independent (right?) P(A and B) = P(A)*P(B). Than the P(A)'s cancel just leaving P(B).

For 2) The part of EXACTLY 3 dice show 2. We know that 3 dice have to show 2 out of the 4, hence the prob. for the first die to be 2 has to equal 3/4. But than for at LEAST 3 dice show 2, I am confused as where to start.

Last edited: Oct 18, 2012
4. Oct 18, 2012

HallsofIvy

Staff Emeritus
Did you try enumerating? There is a 1/4 probability that the two coins show NO heads but the die cannot show "0". There is a 1/2 probability that the two coins show one head and 1/6 probability that the die shows "1". There is a 1/4 probability that the two coins show two heads and 1/6 probability that the die shows "2".

Again, enumeration. The probability that all 4 dice show 2 is $(1/6)^4$. The probability of "at least 3" is $\begin{bmatrix}4\end{bmatrix}(1/6)^3(5/6)+(1/6)^4$.

5. Oct 18, 2012

Ray Vickson

No. If A and B are as you have defined them, you are asked to find Pr{A = B}. The event {A=B} consists of number of disjoint outcomes. What are these outcomes? How would you compute the probabilities of the individual outcomes?

RGV

6. Oct 18, 2012

JP16

As HallsofIvy said the possible outcomes, the set would contain {TH1,HT1, HH2}. Then the probability of that would just be,

2($\frac{2}{4}\frac{1}{6})+\frac{1}{4}\frac{1}{6} = \frac{5}{24}$

Would this be correct?

OH, that makes sense. Then,

P(1st die 2 | least 3 dice show 2) =$\frac{ P (1st \,die\, 2\,\cap\, least\, 3\, dice\, show \,2)}{P(least \,3 \,dice \,show \,2)}$

P(1st die 2) = $\frac{1}{6}(\frac{5}{6})^3$

What would be the relationship(intersection) of A and B? What I think, at least 3 dice showing 2 intersected with the 1st die showing 2 is just the 1st die showing 2? This doesn't completely make sense to me. I have a exam tomorrow, and conditional probability is just being hard on me. So all the help is really appreciated. Thank you.

EDIT: By the way, the answer for #2 is given in the book. It says 16/21.

Last edited: Oct 18, 2012
7. Oct 18, 2012

Ray Vickson

For P{A=B} you write
$$2 \left( \frac{2}{4} \frac{1}{6} \right) + \frac{1}{4} \frac{1}{6}.$$
Where does the (2/4)*(1/6) come from? Where does the 2*(2/4)*(1/6) come from?

RGV

8. Oct 18, 2012

JP16

Oh my bad, it should just be (2/4) * (1/6) + (1/4)*(1/6).
(2/4) : probability of having 2 heads in the total # of outcomes of flipping 2 coins.
(1/6) : probability of rolling a 1.

(1/4) : probability of both coins being heads.
(1/6) : probability of rolling a 2.

Am I making any progress? I feel like I am making it more difficult than it is.

9. Oct 18, 2012

Ray Vickson

This is OK now. Don't despair: these things get easier with practice.

RGV

10. Oct 18, 2012

JP16

Thanks for the fast responses and guiding me in the right direction. Any guidance for the other question?