Question

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.

Short answer

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

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

Step-by-step solution

Step 1:Given Informaiton

Number of trials, $n=6$

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

Step 2:Simplification

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:

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

Step 3:Simplification

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.