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The total number of trials

The likelihood of success$\mathrm{p}=20.5\mathrm{\%}=0.205$

The following are four binomial setup conditions:

• (Success/failure) binary

• Trials that are conducted independently

• The number of trials is fixed.

• The likelihood of success (same for each trial)

The condition has been satisfied because success results in orange candy and failure results in candy that is not tinted orange.

Trials conducted independently: Because the random sampling of 8 candies represents less than 10% of the total population of sweets, As a result, the 10 % condition is safe to conclude that the trials are independent.

The number of trials is fixed because we chose eight sweets and the number of trials is equal eight. As a result, the criterion has been met.

Probability of success: Because the candy has a $\$20.5$percent chance of being orange, the probability of success is also $\$20.5$percent. As a result, the criterion has been met.

The total number of trials $\mathrm{n}=8$

The likelihood of success is $\mathrm{p}=20.5\mathrm{\%}=0.205$

According to the definition of binomial probability,

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}\mathrm{n}\\ \mathrm{k}\end{array}\right)鈰�{\mathrm{p}}^{\mathrm{k}}鈰�(1鈭�\mathrm{p}{)}^{\mathrm{n}鈭�\mathrm{k}}$

For , $\mathrm{k}=3$

The binomial probability is calculated as follows:

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=3\right)=\left(\begin{array}{l}8\\ 3\end{array}\right)鈰�{\left(0.205\right)}^3鈰�{\left(1鈭�0.205\right)}^{8鈭�3}\\ =\frac{8!}{3!(8鈭�3)!}鈰�{\left(0.205\right)}^3鈰�{\left(0.795\right)}^5\\ =56鈰�{\left(0.205\right)}^3鈰�{\left(0.795\right)}^5\\ 鈮�0.1532\end{array}$

Furthermore,

The chance of receiving three orange M&Ms is approximately 0.1532.

The given probability is$\mathrm{P}(\mathrm{X}鈮�4)=0.0610$

The total number of trials $\mathrm{n}=8$

The likelihood of success is $\text{p=20.5\%=0.205}$

According to the definition of binomial probability,

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}\mathrm{n}\\ \mathrm{k}\end{array}\right)鈰�{\mathrm{p}}^{\mathrm{k}}鈰�(1鈭�\mathrm{p}{)}^{\mathrm{n}鈭�\mathrm{k}}$

Mutually exclusive event addition rule:

role="math" $\mathrm{P}\left(\mathrm{A}鈭�\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\text{or}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)$

For $\mathrm{k}=4$,

The binomial probability is calculated as follows:

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=4\right)=\left(\begin{array}{c}8\\ 4\end{array}\right).{\left(0.205\right)}^4.{\left(1鈭�0.205\right)}^{8鈭�4}\\ =\frac{8!}{4!(8鈭�4)!}.{\left(0.205\right)}^4.{\left(0.795\right)}^4\\ 鈮�0.0494\end{array}$

For $\mathrm{k}=5,$

The binomial probability is calculated as follows:

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=5\right)=\left(\begin{array}{c}8\\ 5\end{array}\right).{\left(0.205\right)}^5.{\left(1鈭�0.205\right)}^{8鈭�5}\\ =\frac{8!}{5!(8鈭�5)!}.{\left(0.205\right)}^5.{\left(0.795\right)}^3\\ 鈮�0.0102\end{array}$

For $\mathrm{k}=6,$

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=6\right)=\left(\begin{array}{c}8\\ 6\end{array}\right).{\left(0.205\right)}^6.{\left(1鈭�0.205\right)}^{8鈭�6}\\ =\frac{8!}{6!(8鈭�6)!}.{\left(0.205\right)}^6.{\left(0.795\right)}^2\\ 鈮�0.0013\end{array}$

For $\mathrm{k}=7,$

The binomial probability is calculated as follows:

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=7\right)=\left(\begin{array}{c}8\\ 7\end{array}\right).{\left(0.205\right)}^7.{\left(1鈭�0.205\right)}^{8鈭�7}\\ =\frac{8!}{7!(8鈭�7)!}.{\left(0.205\right)}^7.{\left(0.795\right)}^1\\ 鈮�0.0001\end{array}$

For $\mathrm{k}=8,$

The binomial probability is calculated as follows:

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=8\right)=\left(\begin{array}{c}8\\ 8\end{array}\right).{\left(0.205\right)}^8.{\left(1鈭�0.205\right)}^{8鈭�8}\\ =\frac{8!}{8!(8鈭�8)!}.{\left(0.205\right)}^8.{\left(0.795\right)}^0\\ 鈮�0.0000\end{array}$

Because it's impossible to have two distinct counts of successes in the same simulation.

For mutually exclusive events, use the addition rule:

$$\begin{gathered} \mathrm{P}(\mathrm{X}鈮�4)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7)+\mathrm{P}(\mathrm{X}=8) \\ =0.0494+0.0102+0.0013+0.0001+0.0000 \\ =0.0610 \\ =6.10\% \end{gathered}$$

As a result, we acquire at least 4 orange candies out of a total of 8 candies about 6.10% of the time.

The total number of trials $\mathrm{n}=8$

The likelihood of success is $\mathrm{p}=20.5\mathrm{\%}=0.205$

We have , as a result of Part (c),

$\mathrm{P}(\mathrm{X}鈮�4)=0.0610$

It is deemed modest when the probability is less than 0.05.

However, because 0.0610 is more than 0.05, the likelihood is high in this scenario.

This means there's a good chance you'll get four or more orange candies.

As a result, there is no compelling proof that Mars' M&M claim is untrue.