91Ó°ÊÓ

Number of trials, $n=30$

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

The binomial probability states that

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

Mutually exclusive event addition rule:

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

At $k=0$,

The binomial probability is to be calculated as follows:

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

At $k=1$,

The binomial probability is to be calculated as follows:

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

At $k=2$,

The binomial probability is to be calculated as follows:

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

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

For mutually exclusive events, use the addition rule:

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

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.