91Ó°ÊÓ

Given information.

The probability of the light turning red on a random workday is $55$percent $=0.55$.

The total number of working days is ten.

Let $Y$represent the number of times the light turns red.

Because the variable$Y$satisfies the following properties, it follows a binomial distribution.

1. The number of independent trials

2. Each trial has an equal chance of success or failure.

3. The outcome of each trial is either success or failure.

4. The number of trials is predetermined.

The probability function of $Y$is $P\left(y\right)=\left(\begin{array}{c}10\\ y\end{array}\right)(0.55{)}^y(1-0.55{)}^{10-y}$

Insert the values of $y=0,1,2,…,12$in the probability function to get the individual probabilities.

The probability of no red light on $x=0$is,

$\begin{array}{rcl}P\left(0\right)& =& \left(\begin{array}{l}10\\ 0\end{array}\right)(0.55{)}^0(0.45{)}^{10-0}\\ & =& 0.00034\end{array}$

The possibility of a one-day red light on$x=1$is,

$$\begin{gathered} P\left(1\right)=\left(\begin{array}{l}10\\ 1\end{array}\right)(0.55{)}^1(0.45{)}^{10-1} \\ =120(0.55)(0.45{)}^8 \\ =0.00416 \end{gathered}$$

The probability that two days red light on $x=2$is,

$\begin{array}{rcl}P\left(2\right)& =& \left(\begin{array}{c}10\\ 2\end{array}\right)(0.55{)}^2(0.45{)}^{10-2}\\ & =& \frac{12×11}{2×1}(0.55{)}^2(0.45{)}^8\\ & =& 0.0229\end{array}$

Continue in the same manner to obtain the probabilities.

The probability distribution of the variable Y is shown below.

The probability distribution graph is shown below.

The distribution appears to be roughly symmetric. The greatest likelihood occurs when $x=6$.