Dice role: What is the probability to observe 2 times 1 and 2 times 5 with the outcome of a fifth die role being unknown?

Question

I tried to solve the following exercise:

Given a dice with $P(X=2) = P(X=4) = P(X=5) = \frac{2}{15}$ and $P(X=1) = P(X=6) = P(X=3) = \frac{2}{10}$. What is the probability to observe 2 times 1 and 2 times 5 with the outcome of a fifth die role being unknown? Tip: Use marginalization!

I tried to use the multinomial probability mass function and marginalize: So I got:

\begin{equation*} \begin{split} &\sum^{6}_{i=1} \frac{5}{2! \cdot 2! \cdot x_i!} \cdot (\frac{2}{10})^2 \cdot (\frac{2}{15})^2 \cdot p(x_i)^1\\ =& 0.0213 \end{split} \end{equation*}

and as a result obtained that the probability is 2,13%.

But the real solution should be 1,6%. Can you give me an advice what I did wrong?

Answer 1

You were mostly okay, except for the cases where you have three ones or three fives.   (Also note the multinomial coefficient in the other cases .)

$$\begin{equation*} \begin{split} &\sum_{i\in\{2,3,4,6\}} \tfrac{5!}{2! \cdot 2! \cdot 1!}~ (\tfrac{2}{10})^2~(\tfrac{2}{15})^2 ~\mathsf P(X{=}i)^1 ~+~\tfrac{5!}{3!\cdot 2!}~(\tfrac{2}{10})^3~(\tfrac{2}{15})^2+\tfrac{5!}{3!\cdot 2!}~(\tfrac{2}{10})^2~(\tfrac{2}{15})^3 \\[1ex] =& \tfrac{5!}{2! \cdot 2! \cdot 1!} ~ (\tfrac{2}{10})^2 ~ (\tfrac{2}{15})^2 ~(\tfrac{2}{15}+\tfrac{2}{10}+\tfrac{2}{15}+\tfrac{2}{10}) ~+~\tfrac{5!}{3!\cdot 2!}~(\tfrac{2}{10})^2~(\tfrac{2}{15})^2~(\tfrac{2}{10}+\tfrac{2}{15})\\[1ex] =& 0.016\dot{\overline{592}} \end{split} \end{equation*}$$

Answer 2

I can't tell you what you did wrong, because I'm not sure what you did right.

In any case, split it into disjoint events, and then add up their probabilities.

---

The probability of exactly $2$ ones and $2$ fives:

$$\frac{5!}{2!\cdot2!\cdot(5-2-2)!}\cdot\left(\frac{2}{10}\right)^2\cdot\left(\frac{2}{15}\right)^2\cdot\left(1-\frac{2}{10}-\frac{2}{15}\right)^{5-2-2}$$

---

The probability of exactly $2$ ones and $3$ fives:

$$\frac{5!}{2!\cdot3!}\cdot\left(\frac{2}{10}\right)^2\cdot\left(\frac{2}{15}\right)^3$$

---

The probability of exactly $3$ ones and $2$ fives:

$$\frac{5!}{3!\cdot2!}\cdot\left(\frac{2}{10}\right)^3\cdot\left(\frac{2}{15}\right)^2$$