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Total number of trials,$\mathrm{n}=7$

The likelihood of success, $\mathrm{p}=44\%=0.44$

The binomial probability states that

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

Mutually exclusive event addition rule:

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

For$\mathrm{k}=5$,

The binomial probability is calculated as follows:

$\begin{array}{r}\mathrm{P}(\mathrm{X}=5)=\left(\begin{array}{c}7\\ 5\end{array}\right)鈰�(0.44{)}^5鈰�(1鈭�0.44{)}^{7鈭�5}\\ =\frac{7!}{5!(7鈭�5)!}鈰�(0.44{)}^5鈰�(0.56{)}^2\\ =21鈰�(0.44{)}^5鈰�(0.56{)}^2\\ 鈮�0.1086\end{array}$

For $k=6$,

The binomial probability is calculated as follows:

$\begin{array}{r}\mathrm{P}(\mathrm{X}=6)=\left(\begin{array}{c}7\\ 6\end{array}\right)鈰�(0.44{)}^6鈰�(1鈭�0.44{)}^{7鈭�6}\\ =\frac{7!}{6!(7鈭�6)!}鈰�(0.44{)}^6鈰�(0.56{)}^1\\ =7鈰�(0.44{)}^6鈰�(0.56{)}^1\\ 鈮�0.0284\end{array}$

For K=7,

The binomial probability is calculated as follows:
$\begin{array}{r}\mathrm{P}(\mathrm{X}=7)=\left(\begin{array}{c}7\\ 7\end{array}\right)鈰�(0.44{)}^7鈰�(1鈭�0.44{)}^{7鈭�7}\\ =\frac{7!}{7!(7鈭�7)!}鈰�(0.44{)}^7鈰�(0.56{)}^0\\ =1鈰�(0.44{)}^7鈰�(0.56{)}^0\\ 鈮�0.0032\end{array}$

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

For mutually exclusive events, use the addition rule:

$\begin{array}{r}\mathrm{P}(\mathrm{X}>4)=\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7)\\ =0.1086+0.0284+0.0032\\ =0.1402\end{array}$

Probabilities of less than 0.05 are deemed small.

However, in this scenario, the likelihood is not insignificant.

This indicates that the occurrence is most likely to happen by chance.

As a result, the fact that more than four elk have survived to adulthood is not remarkable.