First I made a mistake of not taking into account that this event is dependent.

So to get two '$6$' the probability would be:

$$ P_5(2) = \frac{5!}{2!(5-2)!} \cdot \left(\frac16\right)^2\cdot\left(\frac56\right)^3 $$

Where

  • $p = \dfrac16$
  • $q = \dfrac56$
  • $N= 5$ (the number of elements of the system)

Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:

$$ P_3(1) = \frac{3!}{1!(3-1)!} \cdot \left(\frac16\right)^1\cdot\left(\frac56\right)^2 $$

But then the professor told me that $p=\dfrac15$ and not $p=\dfrac16$ That's what I don't get.

Of course at the end we multiply those two probabilities to get the final probability that we want.

  • 2
    The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s. – StatGuy Oct 12 at 4:50
up vote 5 down vote accepted

I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.

Method 1

You can solve this using the joint pmf for a multinomial distribution.

$$P(X_6 = 2, X_5 = 1, X_{other} = 2) = \frac{5!}{2!1!2!} \left( \frac{1}{6} \right)^2 \left( \frac{1}{6} \right) \left( \frac{4}{6} \right)^2 \approx .062$$


Method 2

Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.

Let $E$ be the event you get two sixes.

Let $F$ be the event you get one five.

Let $G$ be the event you get two of the others.

$$P(EFG) = P(E)P(F \mid E)P(G \mid EF) $$

$$P(E) = {5 \choose 2}\left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^3 $$

$$P(F \mid E) = {3 \choose 1}\left( \frac{1}{5} \right) \left( \frac{4}{5} \right)^2$$

$$P(G \mid EF) = {2 \choose 2}\left( \frac{4}{4} \right)^2 \left( \frac{0}{4} \right)^0 = 1$$

$$P(EFG) \approx .062$$

  • The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces. – Nuz Oct 12 at 5:14
  • In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space. – HJ_beginner Oct 12 at 5:38
  • Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something? – Nuz Oct 12 at 7:11
  • Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F \mid G)P(E \mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting. – HJ_beginner Oct 12 at 7:36

But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.

You have correctly evaluated the probability for the event that two from the five dice show 6.

$$\mathsf P(N_6{=}2)=\binom 52 \dfrac{1^25^3}{6^5}$$

Now, of the three dice which don't show 6, each may show faces $\{1,2,3,4,5\}$ with equal probability.   Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$.   So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:

$$\mathsf P(N_5{=}1\mid N_6{=}2)=\binom 31\dfrac{1^14^2}{5^3}$$

Of course at the end we multiply those two probabilities to get the final probability that we want.

And the $5^3$ factors will cancel, so:

$$\mathsf P(N_6{=}2,N_5{=}1)=\binom{5}{2}\binom{3}{1}\dfrac{1^21^14^2}{6^5}$$

PS: This is called a multinomial distribution.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.