This problem aims to build our understanding of **random events** and their **predictable outputs.** The concepts behind this problem are primarily associated with **a probability** and **probability distribution.**

We can define **probability** as a way to indicate the **occurrence** of an **unanticipated event,** and the probability can be between **zero** and **one.** It estimates the possibility of an **event,** such events that are difficult to forecast an **output.** Its standard description is that a **likelihood **of an event occurring is equal to the **ratio** of fair outcomes and the total **number** of **trials.**

**Given as:**

\[P(\text{Event to occur})=\dfrac{\text{Favourable Events}}{\text{Total Events}}\]

## Expert Answer

As per the given **statement,** we have $8$ **white,** $4$ **black,** and $2$ **orange balls.** Each **selection** of a **randomly chosen ball** results in a win or a loose denoted b $(X)$. The **possible results** of the **experiment** are:

\[\{WW\},\space \{WO\},\space \{OO\},\space \{WB\},\space \{BO\},\space \{BB\}\]

The values of $(X)$ **corresponding** to the **outcomes** of the **events listed** are:

\[\{WW=-2\},\space \{WO=-1\},\space \{OO=0\},\space \{WB=1\},\space \{BO=2\},\space \{BB=4\}\]

Where $W$ stand for **White,** $O$ for **orange,** and $B$ stands for the **black** ball.

We are to **choose** $2$ **balls** at **random** from a total of $8+4+2 = 14$ **balls,** so the **combination** becomes:

\[C^{n}_{r}=\dfrac{n!}{r!(n-r)!}\]

\[C^{14}_{2}=\dfrac{14!}{2!(14-2)!}\]

\[C^{14}_{2}=\dfrac{14!}{2!\cdot 12!}\]

\[C^{14}_{2}=91\]

The **probability** of **choosing two white balls** is:

\[P(X = -2)=P(\{W, W\})=\dfrac{\begin{pmatrix} 8 \\ 2 \end{pmatrix}}{\begin{pmatrix} 14 \\ 2 \end{pmatrix}}=\dfrac{28}{91} \]

Similarly, the **rest** of the **probabilities** can be **calculated** as follows:

\[P(X = -1)=P(\{W, O\})=\dfrac{\begin{pmatrix} 8 \\ 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix}}{\begin{pmatrix} 14 \\ 2 \end{pmatrix}} = \dfrac{16}{91} \]

\[P(X = 1)=P(\{W, B\})=\dfrac{\begin{pmatrix} 8 \\ 1 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix}}{\begin{pmatrix} 14 \\ 2 \end{pmatrix}}=\dfrac{32}{91} \]

\[P(X = 0)=P(\{O, O\})=\dfrac{\begin{pmatrix} 2 \\ 2 \end{pmatrix}}{\begin{pmatrix} 14 \\ 2 \end{pmatrix}}=\dfrac{1}{91} \]

\[P(X = 2)=P(\{O, B\})=\dfrac{\begin{pmatrix} 2 \\ 1 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix}}{\begin{pmatrix} 14 \\ 2 \end{pmatrix}}=\dfrac{8}{91} \]

\[P(X = 4)=P(\{B, B\}) = \dfrac{\begin{pmatrix} 4 \\ 2 \end{pmatrix}}{\begin{pmatrix} 14 \\ 2 \end{pmatrix}}=\dfrac{6}{91} \]

Since we have the **probability distribution,** we are going to use the **formula** $\mu = \sum x_{\iota} P(X=x_{\iota})$ to find the expected value of $X$:

\[\mu=-2\cdot\dfrac{28}{91}-1\cdot\dfrac{16}{91}+0\cdot\dfrac{1}{91}+1\cdot \dfrac{32}{91}+2\cdot\dfrac{8}{91}+4\cdot\dfrac{6}{91}\]

\[\mu=0\]

## Numerical Result

The **probabilities associated** with each **value** of $X$ are given in the **table:**

## Example

A **claim suffered** that $60\%$ of all solar systems **installed,** the utility bill is decreased by at most **one-third.** Therefore, what could be the **probability** that the utility bill will be **lowered** by at **minimum one-third** in at least **four** out of the **five inductions?**

Assume $X$ be **equal** to **measuring** the number of **reduced utility bills** by at least **one-third** in five **solar systems installations,** with some certain **parameters** $n = 5$, $p = 0.6$ and $q = 1− p = 0.4$. We are **requested** to find the **subsequent probabilities:**

**Part a:**

\[P(X=4)=\begin{pmatrix} 5 \\4\end{pmatrix} (0.6)^4(0.4)^{5−4} = 0.259 \]

**Part b:**

\[P(X\geq 4)=P(X = 4) + P(X = 5) = 0.259+\begin{pmatrix} 5 \\ 5 \end{pmatrix}(0.6)^5 (0.4)^{5−5} = 0.259 + 0.078 = 0.337\]

*Image/Mathematical drawings are created in Geogebra.*