Question

There are n dice with f faces marked from 1 to f. If these are thrown at random, what is the chance that the sum of the numbers exhibited shall be equal to p?

Answer 1

Hint:To solve the question we have to calculate the number of possible outcomes when the n dice with f faces marked from 1 to f, are thrown randomly. To solve further, calculate the number of ways the sum of numbers exhibited when the n dice thrown randomly be equal to p. To calculate the chance the sum of numbers exhibited shall be equal to p, calculate the probability of the sum of numbers exhibited to be equal to p, for the obtained number of possible outcomes when the n dice are thrown randomly.

Complete step-by-step answer:
Given
The number of dices = n
The number of faces in each dice = f
We know,
Number of possible outcomes when dices are randomly thrown =${Number of faces}^{Number of dices}$.
Thus, the number of possible outcomes from the given information of dices \[={{f}^{n}}\]
The sum of numbers exhibited when dices are thrown randomly \[=({{x}^{1}}+{{x}^{2}}+.....+{{x}^{f}}).({{x}^{1}}+{{x}^{2}}+.....+{{x}^{f}})......({{x}^{1}}+{{x}^{2}}+.....+{{x}^{f}})\] \[n\] times. ( \[\because n\] dices)
\[={{({{x}^{1}}+{{x}^{2}}+.....+{{x}^{f}})}^{n}}\]
The number of ways the sum of numbers exhibited shall be equal to \[p\] when dices are thrown randomly = Coefficient of \[{{x}^{p}}\] in the expansion \[{{({{x}^{1}}+{{x}^{2}}+.....+{{x}^{f}})}^{n}}\]
We know,
The formula for the sum of geometric progression \[{{a}^{1}}+{{a}^{2}}+.....+{{a}^{f}}\]is given by \[=\dfrac{a(1-{{a}^{f}})}{1-a}\]
\[\Rightarrow {{({{x}^{1}}+{{x}^{2}}+.....+{{x}^{f}})}^{n}}={{\left( \dfrac{x(1-{{x}^{f}})}{1-x} \right)}^{n}}\]
\[=\dfrac{{{x}^{n}}{{(1-{{x}^{f}})}^{n}}}{{{(1-x)}^{n}}}\]
\[={{x}^{n}}{{(1-{{x}^{f}})}^{n}}{{(1-x)}^{-n}}\]
\[\Rightarrow \] Coefficient of \[{{x}^{p}}=\] Coefficient of \[{{x}^{p-n}}\] in \[{{(1-{{x}^{f}})}^{n}}{{(1-x)}^{-n}}\]
We know the expansion formulae are
\[{{(1-x)}^{n}}=1-nx+\dfrac{n(n-1)}{2!}{{x}^{2}}-........+{{(-1)}^{r}}\dfrac{n(n-1)(n-2)....(n-r+1)}{r!}{{x}^{r}}+.....\infty \]
\[{{(1-x)}^{-n}}=1+nx+\dfrac{n(n+1)}{2!}{{x}^{2}}+........+\dfrac{n(n+1)(n+2)....(n+r-1)}{r!}{{x}^{r}}+.....\infty \] ……. (1)
\[\Rightarrow {{(1-{{x}^{f}})}^{n}}=1-n{{x}^{f}}+\dfrac{n(n-1)}{2!}{{x}^{2f}}-........+{{(-1)}^{r}}\dfrac{n(n-1)(n-2)....(n-r+1)}{r!}{{x}^{rf}}+.....\infty \] ……. (2)

By multiplying (1) and (2) we get the coefficient of \[{{x}^{p-n}}\]
Coefficient of \[{{x}^{r}}\] in \[{{(1-x)}^{-n}}\]\[=\dfrac{n(n+1)(n+2)....(n+r-1)}{r!}\]
Coefficient of \[{{x}^{p-n}}\] in \[{{(1-x)}^{-n}}\] \[=\dfrac{n(n+1)(n+2)....(p-1)}{(p-n)!}\]
Thus, the coefficient of \[{{x}^{p-n}}\] in the product of (1) and (2)
\[=\dfrac{n(n+1)(n+2)....(p-1)}{(p-n)!}-n.\dfrac{n(n-1)(n-2)....(p-f+1)}{(p-n-f)!}+...\]
\[..........+{{(-1)}^{r}}\dfrac{n(n-1)(n-2)....(n-r+1)}{r!}.\dfrac{n(n-1)(n-2)....(p-rf-1)}{(p-n-rf)!}+...\infty \]
The chance the sum of numbers exhibited shall be equal to \[p=\]\[\dfrac{Coefficientof{{x}^{p}}}{{{f}^{n}}}\]
∴ The required probability of the sum of numbers exhibited be equal to \[p\]
\[=\dfrac{\dfrac{n(n+1)(n+2)....(p-1)}{(p-n)!}-n.\dfrac{n(n-1)(n-2)....(p-f+1)}{(p-n-f)!}+...+{{(-1)}^{r}}\dfrac{n(n-1)(n-2)....(n-r+1)}{r!}.\dfrac{n(n-1)(n-2)....(p-rf-1)}{(p-n-rf)!}+...\infty }{{{f}^{n}}}\]

Note: To solve the question we have to apply the sum of geometric progression formulas. We have to analyze that the sum of numbers exhibited is equal to the coefficient of \[{{x}^{p}}\]. While solving after a certain step, find the coefficient of \[{{x}^{p-n}}\] instead of coefficient \[{{x}^{p}}\]which will ease the procedure of solving.