91影视

Number of trials, $n=22$

Probability of success,$p=0.20$

According to the binomial probability,

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

Addition rule for mutually exclusive event:

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

At $k=14$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=14)=\left(\begin{array}{c}22\\ 14\end{array}\right)路(0.20{)}^{14}路(1-0.20{)}^{22-14} \\ =\frac{22!}{14!(22-14)!}路(0.20{)}^{14}路(0.80{)}^8 \\ 鈮�0.0000 \end{gathered}$$

At $k=15$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=15)=\left(\begin{array}{c}22\\ 15\end{array}\right)路(0.20{)}^{15}路(1-0.20{)}^{22-15} \\ =\frac{22!}{15!(22-15)!}路(0.20{)}^{15}路(0.80{)}^7 \\ 鈮�0.0000 \end{gathered}$$

At $k=16$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=16)=\left(\begin{array}{c}22\\ 16\end{array}\right)路(0.20{)}^{16}路(1-0.20{)}^{22-16} \\ =\frac{22!}{16!(22-16)!}路(0.20{)}^{16}路(0.80{)}^6 \\ 鈮�0.0000 \end{gathered}$$

At $k=17$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=17)=\left(\begin{array}{c}22\\ 17\end{array}\right)路(0.20{)}^{17}路(1-0.20{)}^{22-17} \\ =\frac{22!}{17!(22-17)!}路(0.20{)}^{17}路(0.80{)}^5 \\ 鈮�0.0000 \end{gathered}$$

At $k=18$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=18)=\left(\begin{array}{c}22\\ 18\end{array}\right)路(0.20{)}^{18}路(1-0.20{)}^{22-18} \\ =\frac{22!}{18!(22-18)!}路(0.20{)}^{18}路(0.80{)}^4 \\ 鈮�0.0000 \end{gathered}$$

At $k=19$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=19)=\left(\begin{array}{c}22\\ 19\end{array}\right)路(0.20{)}^{19}路(1-0.20{)}^{22-19} \\ =\frac{22!}{19!(22-19)!}路(0.20{)}^{19}路(0.80{)}^3 \\ 鈮�0.0000 \end{gathered}$$

At $k=20$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=20)=\left(\begin{array}{c}22\\ 20\end{array}\right)路(0.20{)}^{20}路(1-0.20{)}^{22-20} \\ =\frac{22!}{20!(22-20)!}路(0.20{)}^{20}路(0.80{)}^2 \\ 鈮�0.0000 \end{gathered}$$

At $k=21$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=21)=\left(\begin{array}{c}22\\ 21\end{array}\right)路(0.20{)}^{21}路(1-0.20{)}^{22-21} \\ =\frac{22!}{21!(22-21)!}路(0.20{)}^{21}路(0.80{)}^1 \\ 鈮�0.0000 \end{gathered}$$

At $k=22$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=22)=\left(\begin{array}{l}22\\ 22\end{array}\right)路(0.20{)}^{22}路(1-0.20{)}^{22-22} \\ =\frac{22!}{14!(22-14)!}路(0.20{)}^{22}路(0.80{)}^0 \\ 鈮�0.0000 \end{gathered}$$

Since two different numbers of successes are impossible on same simulation.

Apply addition rule for mutually exclusive events:

$$\begin{gathered} P(X鈮�14)=P(X=14)+P(X=15)+P(X=16)+P(X=17) \\ +P(X=18) \\ +P(X=19)+P(X=20)+P(X=21) \\ +P(X=22)\backslash \backslash =0.0000+0.0000+0.0000+0.0000+0.0000 \\ +0.0000\backslash \backslash +0.0000+0.0000+0.0000 \\ =0.0000 \\ =0.00\% \end{gathered}$$

Number of trials, $n=22$

Probability of success,$p=0.20$

When the probability is less than $0.05$, it is considered to be small.

In this scenario, keep in mind that the likelihood is low enough.

As a result, it seems doubtful that 14 of the 22 students rated the final kiss the highest rating.

This means that there is compelling evidence that the participants prefer the last thing they taste.