Chapter 6.3, Problem 90E

(a)

To determine

Probability for two or fewer students would win a prize if company’s claim is true.

Answer to Problem 90E

Probability for two or fewer students would win a prize is 0.1028.

Explanation of Solution

Given information:

Number of trials, $n=30$

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

Calculations:

According to the binomial probability,

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

Addition rule for mutually exclusive event:

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

At $k=0$ ,

The binomial probability to be evaluated as:

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

At $k=1$ ,

The binomial probability to be evaluated as:

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

At $k=2$ ,

The binomial probability to be evaluated as:

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

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

Apply addition rule for mutually exclusive events:

  $\begin{array}{l}P\left(X\le 2\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\\ =0.0042+0.0253+0.0733\\ =0.1028\end{array}$

Thus,

If company’s claim is true, probability for two or fewer students would win a prize is 0.1028.

(b)

To determine

Whether the result of two cap winners in Grayson’s class gives convincing evidence that the company’s 1 − in − 6 claim is false.

Answer to Problem 90E

No convincing evidence exists for company’s 1 − in − 6 claim is false.

Explanation of Solution

Given information:

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 indicates

It is likely to obtain two winners.

Thus,

There is no convincing evidence that the company’s 1 − in − 6 claim is false.