Chapter 13.3, Problem 2CYP

To determine

To calculate; to find the required probability of tea

Answer to Problem 2CYP

Solution: $\begin{array}{l}A.P\left(ofteabelow2gallons\right)=\frac{1}{4}or25\%\\ B.P\left(ofteabetween2and3gallons\right)=\frac{3}{8}or37.5\%\end{array}$

Explanation of Solution

A.

  $\begin{array}{l}P\left(ofteabelow2gallons\right)=\frac{requiredgallonsoftea}{totalgallonsoftea}\\ =\frac{2}{8}=\frac{1}{4}\end{array}$

B.

  $\begin{array}{l}P\left(ofteabetween2and3gallons\right)=\frac{requiredgallonsoftea}{totalgallonsoftea}\\ =\frac{3}{8}\end{array}$