Red light! Pedro drives the same route to work on Monday through Friday. His route includes one traffic light. According to the local traffic department, there is a $55\%$chance that the light will be red on a randomly selected work day. Suppose we choose 10 of Pedro's work days at random and let $Y=$the number of times that the light is red.

a. Explain why $Y$is a binomial random variable.

b. Find the probability that the light is red on exactly 7 days.

The total number of trials, $n=10$

The likelihood of success,$p=55\%=0.55$

Y: The number of times the light has been turned red.

When it comes to the binomial random variable,

The following are the four conditions:

Binary: Because success resembles a red light and failure like a light that is not red. As a result, the criterion is met.

Trials conducted independently: Pedro's working days are known to be chosen at random. As a result, the criterion has been met.

The number of trials is fixed: Because we selected ten of Pedro's working days. As a result, the criterion has been satisfied and the number of trials has been increased to ten.

The likelihood of success is: Because there is a 55% likelihood that the light will be red. As a result, the chance of the light being red is 55%, and the condition has been met.

As a result, all of the requirements have been met.

A binomial setup is described by Y in this case.

The total number of trials, $n=10$

The likelihood of success, $p=55\%=0.55$

The binomial probability states that

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

For $k=7$,

The binomial probability to be evaluated as:

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

The probability of the light turning red on exactly 7 days is approximately 16.65%, and the probability is approx.0.1665.