# 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?