Taking the train Refer to Exercise 80 . Would you be surprised if the train arrived on time on fewer than 4 days? Calculate an appropriate probability to support your answer.

Number of trials, $n=6$

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

According to the binomial probability,

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

Addition rule for mutually exclusive event:

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

At $k=0$,

The binomial probability to be evaluated as:

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

At $k=1$,

The binomial probability to be evaluated as:

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

At $k=2$,

The binomial probability to be evaluated as:

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

At $k=3$,

The binomial probability to be evaluated as:

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

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

Apply addition rule for mutually exclusive events:

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

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.