# A 4-sided is rolled n times i. How many possible outcomes do we have? ii. For 0 less...

## Question:

A 4-sided is rolled {eq}n {/eq} times

i. How many possible outcomes do we have?

ii. For {eq}0 \leq i \leq n {/eq}, in how many possible outcomes the side {eq}1 {/eq} appear {eq}i {/eq} times?

iii. Prove

{eq}4^n = \sum_{i = 0}^n \begin{pmatrix}n \\ i \end{pmatrix} 3^i = \sum_{i = 0}^n \begin{pmatrix}n \\ i\end{pmatrix}3^{n - i} {/eq}

## Binomial expansion :

Binomial expansion is the property that states that any positive power of the sum of two number can be expressed in the following form.

{eq}{\left( {a + b} \right)^n} = \sum {^n{C_r}} {a^r}{b^{\left( {n - r} \right)}} {/eq}

1)

Possible outcome if one dice is rolled is 4

Possible outcome if n dice is rolled is

{eq}\begin{align*} pos\;outcomes = 4 \times 4 \times 4 \times 4 \times {......._{\left( {n\;times} \right)}}\\ = {4^n} \end{align*} {/eq}

2)

X follows binomial distribution defining 1 appears n times.

Probability that 1 appear is 0.25.

Therefore, its probability distribution can be defined as:

{eq}\begin{align*} f\left( x \right){ = ^n}{C_i}{\left( p \right)^i}{\left( {1 - p} \right)^{n - i}}\\ { = ^n}{C_i}{\left( {0.25} \right)^i}{\left( {0.75} \right)^{n - i}} \end{align*} {/eq}

3)

To prove:

{eq}\begin{align*} L.H.S = \sum\limits_{i = 0}^n {^n{C_i}{3^i}} \\ = \sum\limits_{i = 0}^n {^n{C_i}{3^i}{1^{\left( {n - i} \right)}}} \end{align*} {/eq}

Using binomial expansion

{eq}\begin{align*} R.H.S = {\left( {3 + 1} \right)^n}\\ = {4^n} \end{align*} {/eq}

Also,

{eq}\begin{align*} R.H.S = \sum\limits_{i = 0}^n {^n{C_i}{3^{\left( {n - i} \right)}}} \\ = \sum\limits_{i = 0}^n {^n{C_i}{3^{\left( {n - i} \right)}}{1^i}} \end{align*} {/eq}

Using binomial expansion

{eq}\begin{align*} L.H.S = {\left( {3 + 1} \right)^n}\\ = {4^n} \end{align*} {/eq}

which is equal to LHS

Hence, proved.