Chapter 6.3, Problem 88E

To determine

Interpretation and calculation for the probability of event $Y\ge 7$ .

Answer to Problem 88E

Probability of the event,

  $P\left(Y\ge 7\right)=0.2660$

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.

Calculations:

According to the binomial probability,

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

Addition rule for mutually exclusive event:

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

At $k=7$ ,

The binomial probability to be evaluated as:

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

At $k=8$ ,

The binomial probability to be evaluated as:

  

At $k=9$ ,

The binomial probability to be evaluated as:

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

At

  $k=10$ ,

The binomial probability to be evaluated as:

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

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

Apply addition rule for mutually exclusive events:

  $\begin{array}{l}P\left(X\ge 7\right)=P\left(X=7\right)+P\left(X=8\right)+P\left(X=9\right)+P\left(X=10\right)\\ =0.1665+0.0763+0.0207+0.0025\\ =0.2660\\ =26.60\%\end{array}$

Thus,

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