91Ó°ÊÓ

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

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}\mathrm{n}\\ \mathrm{k}\end{array}\right)â‹\dots {\mathrm{p}}^{\mathrm{k}}â‹\dots (1−\mathrm{p}{)}^{\mathrm{n}−\mathrm{k}}$

For $k=7$,

The binomial probability to be evaluated as:

$\begin{array}{r}\mathrm{P}(\mathrm{X}=7)=\left(\begin{array}{c}10\\ 7\end{array}\right)â‹\dots (0.55{)}^7â‹\dots (1−0.55{)}^{10−7}\\ =\frac{10!}{7!(10−7)!}â‹\dots (0.55{)}^7â‹\dots (0.45{)}^3\\ =120â‹\dots (0.55{)}^7â‹\dots (0.45{)}^3\\ ≈0.1665\\ =16.65\mathrm{\%}\end{array}$

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