An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.

I have tried to solve this question this way:

Total $6P6 = 720$.

1,2 and 3 can be placed in $2$ locations $= 3^2$

4 and 5 can be placed in 1 location $= 2^1$

and 6 can be placed in two locations $= 2^2$

The probability is $= 72/720 = 0.1$

Is this correct?

  • 1
    $\begingroup$ Please read this MathJax tutorial, which explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 14 '18 at 13:00

Close. You are rolling 5 dice, not 6. And you are not looking to permute the numbers, you are looking to find the number of ways you can roll 5 dice. You need to permute the $E_1$ and $E_2$ events. If 6 can be in two locations, then you have $1^2$ ways of placing 6 in 2 locations. Next, you need to permute the multiset:

$$\{E_1\cdot 2, E_2 \cdot 1, 6\cdot 2\}$$

There are


ways to permute this multiset.

So, the answer is:

$$\dfrac{3^2\cdot 2^1\cdot 1^2 \cdot \dfrac{5!}{2!1!2!}}{6^5} = \dfrac{5}{72}$$

  • $\begingroup$ Can you please explain why two terms are being multiplied in numerator of last value ? $\endgroup$ – puffles Sep 14 '18 at 13:29
  • $\begingroup$ I will use 1 to mean a value of 1, 2, or 3 on the die and 4 to mean a value of 4 or 5 on the die. So, 11466 would be an order of the dice that are rolled where the first two dice show numbers between 1 and 3, the third shows either 4 or 5, and the last two are both 6. This is a valid outcome. Other valid outcomes are: 11646, 11664, 14166,... The term $$\dfrac{5!}{2!1!2!}$$ is the number of ways to order the digits 11466. $\endgroup$ – InterstellarProbe Sep 14 '18 at 13:45
  • $\begingroup$ And why is it being multiplied with 3^2*2^1*1^2? I hope I'm not bothering too much :D I just need to get my concepts cleared $\endgroup$ – puffles Sep 14 '18 at 13:47
  • $\begingroup$ How many ways can you get 11466? That would be $3^2\cdot 2^1\cdot 1^2$. How many ways can you get 14166? That would be $3\cdot 2\cdot 3\cdot 1^2 = 3^2\cdot 2^1\cdot 1^2$. Multiplication is commutative. So, for each order, the number of ways to achieve the die rolls remains constant. So, we multiply the number of ways to achieve a single order by the number of possible orders. $\endgroup$ – InterstellarProbe Sep 14 '18 at 15:48

The probability equals:$$\binom5{2,1,2}\left(\frac36\right)^2\left(\frac26\right)^1\left(\frac16\right)^2=\frac5{72}$$(where $\binom5{2,1,2}:=\frac{5!}{2!1!2!}$)

You can look at this as a case of trinomial distribution.

There are $5$ independent experiments and there are not $2$ possible and mutually exclusive outcomes (success and failure, as is the case by binomial distribution) but there are $3$ possible and mutually exclusive outcomes: the die shows a face in $\{1,2,3\}$, or in $\{4,5\}$ or in $\{6\}$. This with probabilities $\frac36$, $\frac26$ and $\frac16$ that sum up to $1$.


There are six possible outcomes for each of the five throws, so your denominator should be $6^5$.

For the favorable cases, we must have a sequence of five throws in which three of the outcomes are 1, 2, or 3, two of the outcomes are 4 or 5, and two of the outcomes are a six. There are $\binom{5}{2}$ ways to select which two positions in the sequence are occupied by a 1, 2, or 3 and $3$ ways to fill each of those positions. That leaves three positions to fill. There are $\binom{3}{1}$ ways to choose which of these positions will be occupied by a 4 or 5 and two ways to fill that position. The remaining two positions of the sequence must be filled with a 6. Therefore, the number of favorable cases is $$\binom{5}{2}3^2 \cdot \binom{3}{1}2^1 \cdot \binom{2}{2}1^2 = \binom{5}{2}\binom{3}{1}\binom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2$$

Hence, the probability of obtaining a 1, 2, or 3 exactly twice and a 4 or 5 exactly once in five rolls of a fair die is $$\frac{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5}$$

It is illustrative to compare this result with those provided by drhab and InterstellarProbe.

Observe that $$\binom{5}{2}\binom{3}{1}\binom{2}{2} = \frac{5!}{2!3!} \cdot \frac{3!}{1!2!} \cdot \frac{2!}{2!0!} = \frac{5!}{2!1!2!} = \binom{5}{2, 1, 2}$$ where the multinomial coefficient $$\binom{5}{2,1,2}$$ counts the number of ways of choosing a sequence of length five composed of two objects of one type (in this case, a roll that produces a 1, 2, or 3), one object of a second type (in this case, a roll that produces a 4 or 5), and two objects of a third type (in this case, a roll that produces a 6).

Therefore, \begin{align*} \frac{\dbinom{5}{2}\dbinom{3}{2}\dbinom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5} & = \frac{\dfrac{5!}{2!1!2!} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5}\\ & = \frac{5!}{2!1!2!}\left(\frac{3}{6}\right)^2\left(\frac{2}{6}\right)^1\left(\frac{1}{6}\right)^2\\ & = \binom{5}{2,1,2}\left(\frac{3}{6}\right)^2\left(\frac{2}{6}\right)^1\left(\frac{1}{6}\right)^2 \end{align*}


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