Chapter 11.9, Problem 35PPSE

Solution

(a)

To calculate: The probability that one out of ten boxes falls short of the weight mentioned on it.

The probability that one out of ten boxes falls short of the weight mentioned on it is 0.0914.

Given:

The probability that a cereal box has at least as much weight mentioned on it $p=0.99$

The total number of boxes $n=10$

Concept used:

The probability with respect to a binomial probability distribution for x number of success with parameters $(n,p)$ where n is the total number of trials and p is the probability of success is calculated using the formula $P(X=x)=\frac{n!}{x!\left(n-x\right)!}{p}^x{(1-p)}^{n-x}$

Calculation:

The probability that a randomly chosen box will have at least the weight mentioned on it $p=0.99$ .

The probability that a randomly chosen box will have its weight less than what is mentioned on the box $1-p=0.01$

The total number of boxes picked $n=10$

The probability that there are one out of ten boxes that fall short of the weight mentioned on it is calculated as shown below

The probability that there is one out of 10 that falls below the mentioned weight is equivalent to the probability that there are nine out of ten that have at least the weight mentioned on it.

  $\begin{array}{}$ P(X=9)=\frac{10!}{9!\left(10-9\right)!}{(0.99)}^9{(1-0.99)}^{10-9}$$ P(X=9)=10\times {(0.99)}^9{(0.01)}^1$$ P(X=9)=0.09135$\end{array}$

The probability that there are nine out of ten boxes that have at least the weight mentioned on it is 0.0914(rounded to 4 decimal places).

(b)

To calculate: The probability that there are three boxes out of ten that have less weight than the weight claimed on box and explain.

The probability that there are three boxes out of ten that have less weight than the weight claimed on the box is 0.0001. This is not a usual event to occur as the probability is very low.

Calculation:

The probability that a randomly chosen box will have at least the weight mentioned on it $p=0.99$ .

The probability that a randomly chosen box will have its weight less than what is mentioned on the box $1-p=0.01$

The total number of boxes picked $n=10$

The probability that there are three out of ten boxes that fall short of the weight mentioned on it is calculated as shown below

The probability that there are three out of 10 that fall below the mentioned weight is equivalent to the probability that there are seven out of ten that have at least the weight mentioned on it.

  $\begin{array}{}$ P(X=7)=\frac{10!}{7!\left(10-7\right)!}{(0.99)}^7{(1-0.99)}^{10-7}$$ P(X=7)=120\times {(0.99)}^7{(0.01)}^3$$ P(X=7)=0.00011$\end{array}$

The probability that three out of ten boxes have weight less than that is mentioned on the box is an unusual event as it probability of occurrence is very low $\left(0.0001<0.05\right)$ .