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Suppose we have 4 fair six-sided dice of different colours and faces numbered 1,2,...,6 are rolled independently. (a) How many ways can a total of

i. 4

ii. 5

iii. 6 be obtained?

(b) Compute (to 4 decimal places) the probability of rolling a total of at most 6.

(c) Noting that the rolls are independent, compute (to 4 decimal places) the normal approximation with continuity correction for the probability in the previous part.

Does the "dice of different colours" contribute to anything???

Also, part c has completely lost me...

Help please!

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closed as off-topic by Did, Semiclassical, Omnomnomnom, Travis, Moishe Kohan Sep 24 '14 at 4:47

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  • $\begingroup$ Different colour just means that all dices are considered to be different. I have no idea what part c) means but part a, b is simple counting problem. $\endgroup$ – Jack Yoon Sep 18 '14 at 11:12
  • $\begingroup$ so for a) (i), would it only be one way or 4! ways? (4! to arrange the order the dice are rolled) $\endgroup$ – toiz Sep 18 '14 at 11:25
  • $\begingroup$ Only one way, all dice must show 1. $\endgroup$ – Christoph Sep 18 '14 at 11:29
  • $\begingroup$ hmm, how about a) (ii) ? it would 4 right? $\endgroup$ – toiz Sep 18 '14 at 11:37
  • $\begingroup$ Just as a comment... the number of ways for a sum s with j addends, if you can use all natural numbers from 1 to s, is $\binom{s-1}{j-1}$. In the case for take a sum of 4 with 4 addends is, obviously, $\binom{3}{3}=1$. For $s=5$ is $\binom{4}{3}=4$. And so on. $\endgroup$ – Masacroso Sep 18 '14 at 12:04
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Answering A:

i) $C(X=4)=\dbinom{4}{4}=1$:

  • $(1,1,1,1)$

ii) $C(X=5)=\dbinom{4}{3}=4$:

  • $(1,1,1,2)$
  • $(1,1,2,1)$
  • $(1,2,1,1)$
  • $(2,1,1,1)$

iii) $C(X=6)=\dbinom{4}{3}+\dbinom{4}{2}=10$:

  • $(1,1,1,3)$
  • $(1,1,3,1)$
  • $(1,3,1,1)$
  • $(3,1,1,1)$
  • $(1,1,2,2)$
  • $(1,2,1,2)$
  • $(1,2,2,1)$
  • $(2,1,1,2)$
  • $(2,1,2,1)$
  • $(2,2,1,1)$

Answering B:

The total number of combinations is $6^4$, therefore:

$P(X\leq6)=P(X=4)+P(X=5)+P(X=6)=\dfrac{1}{6^4}+\dfrac{4}{6^4}+\dfrac{10}{6^4}\approx0.0115$

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  • $\begingroup$ how about part b??? Does the fact that the dice are different coloured mean that you have to times everything by 4 factorial ? $\endgroup$ – toiz Sep 18 '14 at 11:52
  • $\begingroup$ @tkbeans: We don't care about colors at that point. Even if they were all the same color, and the size of the sample-space was smaller than $6^4$, for the sake of probability we would still have to take it as $6^4$ (in order to "emulate" samples that appear more than once). $\endgroup$ – barak manos Sep 18 '14 at 11:54
  • $\begingroup$ so when does the colours come into play? Also if it were being rolled independently, shouldn't you also consider the order they were thrown? So times everything by 4! to the # of outcomes??? $\endgroup$ – toiz Sep 18 '14 at 11:55
  • $\begingroup$ Just assume that you are throwing them simultaneously. $\endgroup$ – barak manos Sep 18 '14 at 11:56
  • $\begingroup$ but the q states "rolled independently" edit: don't know if that makes a difference $\endgroup$ – toiz Sep 18 '14 at 11:57
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We deal mainly with part c). For the other two parts, we need to define the sample space precisely. Let the dice be blue, green, red, and yellow. Imagine recording the outcome of the tossing as a sequence of length $4$ made up of symbols $1$ to $6$. The sequence gives the results of the tossing, in colour alphabetical order. So $(5,2,5,1)$ means $5$ on blue, $2$ on green, $5$ on red, $1$ on yellow. Alternately, one can imagine tossing the dice one at a time, and recording the results in time order.

Colouring, or time order, are two ways to visualize concretely the right kind of sample space. This sample space has $6^4$ elements, and all sequences of length $4$ are equally likely.

As has been thoroughly discussed already in comments and an answer, for total $\le 6$ there are $15$ "favourables," so the required probability is $\frac{15}{6^4}$.

Part c): Let $X_1$ be the result on the blue, $X_2$ the result on the green, and so on. Let $S=X_1+\cdots +X_4$. We want $\Pr(S\le 6)$.

Note that $S$ is the sum of $4$ "nice" independent identically distributed random variables. By the Central Limit Theorem, the sum of a large number of nice independent identically distributed random variables has a close to normal distribution.

Now $4$ is definitely not large! But we are asked to use the normal approximation nonetheless.

We need the mean and variance of $S$. Each $X_i$ has mean $\frac{7}{2}$ and variance $\frac{35}{12}$. (We assume that you know how to calculate these.)

It follows that $S$ has mean $14$ and variance $\frac{35}{3}$.

Let $Y$ be normal with mean $14$ and variance $\frac{35}{3}$. The probability that $S\le 6$ will be approximated by the probability that $Y\le 6.5$.

Why the mysterious $0.5$ added to $6$? Because often when we approximate the distribution of a random variable $T$ that takes on only integer values by a normal $W$, $\Pr(T\le n)$ is often better approximated by $\Pr(W\le n+0.5)$ than by $\Pr(W\le n)$. This is the famous continuity correction.

So in our case, $\Pr(S\le 6)\approx \Pr(Y\le 6.5)$ where $Y$ is normal mean $14$ and standard deviation $\sqrt{35/3}$. This is $$\Pr\left(Z\le \frac{6.5-14}{\sqrt{35/3}}\right),$$ where $Z$ is standard normal. And finally we can compute!

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ad C)

If the random variables are independent, then $Var \left (\sum_{i=1}^n X_i \right)=Var(X_1)+Var(X_1)+Var(X_1)+\ldots + Var(X_n)$

Here it is $Var(X_1)=Var(X_2)=Var(X_3)=Var(X_4) \Rightarrow Var \left (\sum_{i=1}^4 X_i \right) =4\cdot \frac{35}{12}$

And $E(\sum_{i=1}^n X_i)=4\cdot 3.5$

The formula for approximation is $P(S_n\leq S)=\Phi \left( \frac{S+0.5-E\left(\sum_{i=1}^n X_i \right)}{\sqrt{Var \left (\sum_{i=1}^4 X_i \right)}} \right)=\Phi \left( \frac{S+0.5-14}{\sqrt{\frac{35}{3}}} \right) $

$\Phi(.)$ is the cdf of standard normal distribution.

$S_n=X_1+X_2+\ldots + X_n$

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