Chapter 13.6, Problem 13E

To determine

To find: the probability of not getting right color from a bag if its use $4$ time in a day.

Answer to Problem 13E

The probability not getting right color out of $4$ times is $\frac{1}{81}$ .

Explanation of Solution

Given:

The probability of getting the right color is $\frac{2}{3}$ .

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

Independent events:

  $p\left(A\cap B\right)=p\left(A\right)\times p\left(B\right).$

Calculation:

The probability of getting the right color is $\frac{2}{3}$ ,then the probability of getting the wrong color is $\frac{1}{3}$ .

So, according to the given not getting the right color out of $4$ times means that it’s multiplied by $\frac{1}{3}$ for each time she gets the wrong color. this give the result as:

According to the given:

  $\begin{array}{l}P\left(\text{ right color}\right)=\frac{2}{3}.\\ P\left(\text{not right color}\right)=\frac{1}{3}.\\ \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{1}{81}.\end{array}$

Hence, the probability not getting right color out of $4$ times is $\frac{1}{81}$ .