91Ó°ÊÓ

Number of trials, $n=10$

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

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

According to the binomial probability,

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

Addition rule for mutually exclusive event:

$P(A∪B)=P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$

At $k=7$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=7)=\left(\begin{array}{c}10\\ 7\end{array}\right)·(0.55{)}^7·(1-0.55{)}^{10-7} \\ =\frac{10!}{7!(10-7)!}·(0.55{)}^7·(0.45{)}^3 \\ =120·(0.55{)}^7·(0.45{)}^3 \\ ≈0.1665 \end{gathered}$$

At $k=8$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=8)=\left(\begin{array}{c}10\\ 8\end{array}\right)·(0.55{)}^8·(1-0.55{)}^{10-8} \\ =\frac{10!}{8!(10-8)!}·(0.55{)}^8·(0.45{)}^2 \\ =45·(0.55{)}^8·(0.45{)}^2 \\ ≈0.0763 \end{gathered}$$

At $k=9$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=9)=\left(\begin{array}{c}10\\ 9\end{array}\right)·(0.55{)}^9·(1-0.55{)}^{10-9} \\ =\frac{10!}{9!(10-9)!}·(0.55{)}^9·(0.45{)}^1 \\ =10·(0.55{)}^9·(0.45{)}^1 \\ ≈0.0207 \end{gathered}$$

At $k=10,$

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=10)=\left(\begin{array}{c}10\\ 10\end{array}\right)·(0.55{)}^{10}·(1-0.45{)}^{10-10} \\ =\frac{10!}{10!(10-10)!}·(0.55{)}^{10}·(0.45{)}^0 \\ =1·(0.55{)}^{10}·(0.45{)}^0 \\ ≈0.0025 \end{gathered}$$

Since two different numbers of successes are impossible on same simulation.

Apply addition rule for mutually exclusive events:

role="math" localid="1653975671376" $$\begin{gathered} P(Xâ‰\yen 7)=P(X=7)+P(X=8)+P(X=9)+P(X=10) \\ =0.1665+0.0763+0.0207+0.0025 \\ =0.2660 \\ =26.60\% \end{gathered}$$

Thus,

Around $26.60\%$chances are there for the Pedro getting at least 7 red lights on the 10 randomly selected working days