Probability that in one rolling of 5 dice we obtain two '6' and one '5'?

Question

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.

Answer 1

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.

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

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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.

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

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

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

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

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

Answer 2

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.