Chapter 6.3, Problem 86E

To determine

Whetherthe arrival of train on time on fewer than 4 days is surprising and calculate an appropriate probability.

Answer to Problem 86E

Yes, the arrival of train on time on fewer than 4 days issurprising and the appropriate probability is 0.0158.

Explanation of Solution

Given information:

Number of trials, $n=6$

Probability of success, $p=90\%=0.90$

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:

  $\begin{array}{l}P\left(X=0\right)=\left(\begin{array}{c}6\\ 0\end{array}\right)\cdot {\left(0.90\right)}^0\cdot {\left(1-0.90\right)}^{6-0}\\ =\frac{6!}{0!\left(6-0\right)!}\cdot {\left(0.90\right)}^0\cdot {\left(0.10\right)}^6\\ =0\cdot {\left(0.90\right)}^0\cdot {\left(0.10\right)}^6\\ \approx 0.0000\end{array}$

At $k=1$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=1\right)=\left(\begin{array}{c}6\\ 1\end{array}\right)\cdot {\left(0.90\right)}^1\cdot {\left(1-0.90\right)}^{6-1}\\ =\frac{6!}{1!\left(6-1\right)!}\cdot {\left(0.90\right)}^1\cdot {\left(0.10\right)}^5\\ =6\cdot {\left(0.90\right)}^1\cdot {\left(0.10\right)}^5\\ \approx 0.0001\end{array}$

At $k=2$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=2\right)=\left(\begin{array}{c}6\\ 2\end{array}\right)\cdot {\left(0.90\right)}^2\cdot {\left(1-0.90\right)}^{6-2}\\ =\frac{6!}{2!\left(6-2\right)!}\cdot {\left(0.90\right)}^2\cdot {\left(0.10\right)}^4\\ =15\cdot {\left(0.90\right)}^2\cdot {\left(0.10\right)}^4\\ \approx 0.0012\end{array}$

At $k=3$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=3\right)=\left(\begin{array}{c}6\\ 3\end{array}\right)\cdot {\left(0.90\right)}^3\cdot {\left(1-0.90\right)}^{6-3}\\ =\frac{6!}{3!\left(6-3\right)!}\cdot {\left(0.90\right)}^3\cdot {\left(0.10\right)}^3\\ =20\cdot {\left(0.90\right)}^3\cdot {\left(0.10\right)}^3\\ \approx 0.0146\end{array}$

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

Apply addition rule for mutually exclusive events:

  $\begin{array}{l}P\left(X<4\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)\\ =0.0000+0.0001+0.0012+0.0146\\ =0.0158\end{array}$

The probability less than 0.05 is considered to be small.

But in this case,

The probability is small.

This implies

The event is unlikely to occur by chance.

Thus,

It is surprising that the train arrives on time on fewer than 4 days.