91影视

Number of trials, $n=30$

Probability of success,$p=\frac{1}{6}$

The binomial probability states that

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

Mutually exclusive event addition rule:

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

At $k=0$,

The binomial probability is to be calculated as follows:

$$\begin{gathered} P(X=0)=\left(\begin{array}{c}30\\ 0\end{array}\right)路(0.44{)}^0路(1-0.44{)}^{30-0} \\ =\frac{30!}{0!(30-0)!}路(0.44{)}^0路(0.56{)}^{30} \\ =1路(0.44{)}^0路(0.56{)}^{30} \\ 鈮�0.0042 \end{gathered}$$

At $k=1$,

The binomial probability is to be calculated as follows:

$$\begin{gathered} P(X=1)=\left(\begin{array}{c}30\\ 1\end{array}\right)路(0.44{)}^1路(1-0.44{)}^{30-1} \\ =\frac{30!}{1!(30-1)!}路(0.44{)}^1路(0.56{)}^{29} \\ =30路(0.44{)}^1路(0.56{)}^{29} \\ 鈮�0.0253 \end{gathered}$$

At $k=2$,

The binomial probability is to be calculated as follows:

$$\begin{gathered} P(X=2)=\left(\begin{array}{c}30\\ 2\end{array}\right)路(0.44{)}^2路(1-0.44{)}^{30-2} \\ =\frac{30!}{2!(30-2)!}路(0.44{)}^2路(0.56{)}^{28} \\ =435路(0.44{)}^2路(0.56{)}^{28} \\ 鈮�0.0733 \end{gathered}$$

Because two distinct counts of successes on the same simulation are impossible.

For mutually exclusive events, use the addition rule:

$$\begin{gathered} P(X鈮�2)=P(X=0)+P(X=1)+P(X=2) \\ =0.0042+0.0253+0.0733 \\ =0.1028 \end{gathered}$$

Thus,

If the company's assertion is correct, the likelihood of two or fewer pupils winning a prize is $0.1028.$

Number of trials, $n=30$

Probability of success,$p=\frac{1}{6}$

The probability less than $0.05$is considered to be small.

Note that

From Part (a),

The probability of at most 2 winners is $0.1028$, which is very much larger than $0.05$

This demonstrates

It seems likely that there will be two winners.

Thus,

There is no compelling evidence that the firm's$1-$in - 6 claim is false.