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**In existing experimentation,**$100$**customers were asked to choose the design they liked. The subsequent data were acquired. Do the data demonstrate the thought that one design is just as conceivable to be designated as another? Explain.**

This problem aims to familiarize us with the concept of the **null hypothesis** and **probability distribution.** The concept of **inferential statistics** is used to explain the **problem,** in which the **null hypothesis** helps us test different **relationships** among different **phenomena.**

In mathematics, the **null hypothesis,** directed to as $H_0$, declares that the **two** occurring **prospects** are **exact.** Whereas the **probability distribution** is a **statistical** procedure that **represents** all the potential **values** and **possibilities** that a spontaneous **variable** can handle within a **provided range.**

## Expert Answer

According to the **given statement,** the **null hypothesis** $H_0$ can be obtained as; all the **designs** are just as **likely** to be **selected** as any **other design,** whereas the **alternative** hypothesis $H_a$ can be **counter positive** of the above **statement,** that is all **designs** are **not given** the **same preference,** then the **probability** of **selecting** a **single package** can be given as:

\[ P(X) = \dfrac{1}{5} = 0.20 \]

But according to the **probability distribution,** we can **achieve** the following results:

The **probability** that the **first** **design** gets chosen is,

\[ P(X = 1) = 0.05 \]

The **probability** that the **second design** gets chosen is,

\[ P(X = 2) = 0.15 \]

The **probability** that the **third design** gets chosen is,

\[ P(X = 3) = 0.30 \]

The **probability** that the **fourth design** gets chosen is,

\[ P(X = 4) = 0.40 \]

The **probability** that the **fifth design** gets chosen is,

\[ P(X = 3) = 0.10 \]

Hence, from the above **probability distribution,** we can notice that the **probability** of choosing any of the **above** $5$ designs is not the **same.**

Thus the **designs** are not just as **equally likely** to each other hence **rejecting** our **null hypothesis.** In order to make the **selection** to be **equally likely,** a **probability** of about $0.20$ would be assigned using the **relative frequency distribution method.**

## Numerical Result

The **probability** of **choosing** any of the given $5$ **designs** is **not** the **same.** Thus, the **designs** are not **just** as **equally likely** to each other, hence it **rejects** the **null hypothesis.**

## Example

**Consider** that a **sample space** has $5$ equally likely **practical results,** $E_1, E_2, E_3, E_4, E_5$, let,

\[ A = [E_1, E_2] \]

\[B = [E_3, E_4] \]

\[C = [E_2, E_3, E_5] \]

Find the **probability** of $A$, $B$, $C$, and $P(AUB)$.

Following are the **probabilities** of $A$, $B$, and $C$:

\[ P(A) = P(E_1, E_2) = \dfrac{2}{5} = 0.4 \]

\[ P(B) = P(E_3, E_4) = \dfrac{2}{5} = 0.4 \]

\[ P(C) = P(E_2, E_3, E_5) = \dfrac{3}{5} = 0.6 \]

**Probability** of $AUB$:

\[ P(AUB) = P(A) + P(B) \]

\[ P(AUB) = P(E_1, E_2) + P(E_3, E_4)\]

\[P(AUB) = P(E_1, E_2, E_3, E_4)\]

\[P(AUB) = \dfrac{4}{5} \]

\[P(AUB) = 0.80 \]