Chapter 10.4, Problem 21E

Solution

a)

To calculate: The probability that there are 5 days in 10 day period, the school bus has to stop at the traffic light..

The probability is 0.2007.

Given:

The percentage that traffic light is red at the driveway leading into the school$\left(p\right)=40\%=0.40$ .

The total number of days $\left(n\right)=10$

Concept used:

A probability distribution where each trial has two possible outcomes known as a success and failure, each trial is independent and the probability of each trial is the same, then to calculate the probability of a certain number of successes occurring, a binomial probability distribution is used as shown below

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

where k is the number of successes, n is the number of trials and p is the probability of success.

Calculation:

Consider, X be the random variable that shows the number of days the bus will stop at the red light is following the binomial distribution.

The probability that the school bus stops at the traffic light for exactly 5 days in 10-day period is calculated as shown below

  $\begin{array}{c}$ P(X=5)=\left(\begin{array}{c}10\\ 5\end{array}\right){(0.4)}^5{(1-0.4)}^{10-5}$=\frac{10!}{5!\left(10-5\right)!}\times {0.4}^5\times {0.6}^5\\ =0.20065\end{array}$

Conclusion:

The required probability is 0.2007 (rounded to 4 decimal places).

b)

To calculate: The probability that the school bus stops at the traffic light at least 8 days in 10 day period.

The probability is 0.0123.

Calculation:

The probability that the school bus stops at the traffic light at least 8 days in 10-day period is calculated as shown below

  $\begin{array}{c}P\left(X\ge 8\right)=P(X=8)+P(X=9)+P(X=10)\\ $ =\left(\begin{array}{c}10\\ 8\end{array}\right){(0.4)}^8{(1-0.4)}^{10-8}+\left(\begin{array}{c}10\\ 9\end{array}\right){(0.4)}^9{(1-0.4)}^{10-9}+\left(\begin{array}{c}10\\ 10\end{array}\right){(0.4)}^{10}{(1-0.4)}^{10-10}$=\frac{10!}{8!\left(10-8\right)!}\times {0.4}^8\times {0.6}^2+\frac{10!}{9!\left(10-9\right)!}\times {0.4}^9\times {0.6}^1+\frac{10!}{10!\left(10-10\right)!}\times {0.4}^{10}\times {0.6}^0\\ =0.012295\end{array}$

Conclusion:

The required probability is 0.0123 (rounded to 4 decimal places).