Chapter 6.3, Problem 84E

(a)

To determine

The number of times the light is red is a binomial random variable.

Answer to Problem 84E

Since all four conditions have been satisfied, Y is a binomial random variable.

Explanation of Solution

Given information:

Number of trials, $n=10$

Probability of success, $p=55\%=0.55$

Y : The number of times that the light is red.

For the binomial random variable,

The four conditions are as follows:

Binary: Since the success resembles the light is red and failure resembles the light is not is not red. Thus, the condition is satisfied.

Independent trials: It is known that Pedro’s working days are chosen at random. Thus, the condition has been satisfied.

Fixed number of trials: Since we chose 10 Pedro’s work days. Thus, the number of trials is 10 and the condition has been satisfied.

Probability of success: Since there are 55% chances for the light to be red. Thus, the probability for the light to be red is 55% and the condition has been satisfied.

Hence,

All the conditions are satisfied.

According to this scenario,

Y describes a binomial setting.

(b)

To determine

Probability for the light is red on exactly 7 days.

Answer to Problem 84E

The probability that the light is red on exactly 7 days is approx. 0.1665.

Explanation of Solution

Given information:

Number of trials, $n=10$

Probability of success, $p=55\%=0.55$

According to the binomial probability,

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

At $k=7$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=7\right)=\left(\begin{array}{c}10\\ 7\end{array}\right)\cdot {\left(0.55\right)}^7\cdot {\left(1-0.55\right)}^{10-7}\\ =\frac{10!}{7!\left(10-7\right)!}\cdot {\left(0.55\right)}^7\cdot {\left(0.45\right)}^3\\ =120\cdot {\left(0.55\right)}^7\cdot {\left(0.45\right)}^3\\ \approx 0.1665\\ =16.65\%\end{array}$

Thus,

Around 16.65% chances are there for the light to be red on exactly 7 days and the probability is approx. 0.1665.