Answer:
A. 0.14
B. 0.0048
C. 0.3467
Explanation:
You are given the table
![\begin{array}{ccccc}\text{Credit Card}&\text{Visa}&\text{Master}&\text{American Express}&\text{Discover}\\ \\\text{East Coast}&0.45&0.37&0.16&0.02\\\text{Midwest}&0.40&0.43&0.03&?\end{array}](https://img.qammunity.org/2020/formulas/mathematics/college/wtl0c78fc1ziu6igfnswe10fhfzrgg5ohe.png)
The sum in each row must be equal to 1, so
![0.40+0.43+0.03+?=1\\ \\?=1-0.86=0.14](https://img.qammunity.org/2020/formulas/mathematics/college/9wtu1gnkdxelklsrwtmtrsgba9twsez289.png)
Hence, the table is
![\begin{array}{ccccc}\text{Credit Card}&\text{Visa}&\text{Master}&\text{American}&\text{Express}\\ \\\text{East Coast}&0.45&0.37&0.16&0.02\\\text{Midwest}&0.40&0.43&0.03&0.14\end{array}](https://img.qammunity.org/2020/formulas/mathematics/college/b2cp529wg5axtwv20diphjutz943agnvoe.png)
A. The probability that a person from Midwest has Discover card is
![0.14](https://img.qammunity.org/2020/formulas/mathematics/college/v48wlv4ib6v4vpyfig6lcbo0sjpj92wfhy.png)
B. The probability that a person chosen at random from East Coast and Midwest, independently of one another, both have type American Express is
![0.16\cdot 0.03=0.0048](https://img.qammunity.org/2020/formulas/mathematics/college/aa0e31wurodmnwtu3tbm995l1jsm9syxm7.png)
C. The probability that a person chosen at random from East Coast and Midwest, independently of one another, both have the same type of credit card is
![0.45\cdot 0.40+0.37\cdot 0.43+0.16\cdot 0.03+0.02\cdot 0.14=0.3467](https://img.qammunity.org/2020/formulas/mathematics/college/y9vfvio089fllztc2qqyzyexk46395ib79.png)