Q:

Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, one chip has the number 3, and the other chip has the number 5. Miguel must choose two chips, and if both chips have the same number, he wins $2. If the two chips he chooses have different numbers, he loses $1 (–$1). Let X = the amount of money Miguel will receive or owe. Fill out the missing values in the table. (Hint: The total possible outcomes are six because there are four chips and you are choosing two of them.)

Accepted Solution

A:
Answer:See explanationStep-by-step explanation:Step 1Miguel wins $2 is if he pulls two chips with the number 1.  Probability of winning is: 2/4 * 1/3 = 1/6  as there are 4 chips in total Probability of loosing is:1- 1/6 = 5/6Step 2Missing values we found in the step 1, can be populated in the table:Xi === 2 === -1P(Xi) === 1/6 === 5/6Expected Value as per table data:1/6*2 + 5/6*(-1) = -1/2 Expected value is $1/2 loss each time he plays Step 3To make the game fair, the expected value should be zero.Then as per the calculation above, let's replace 2, with x, and find its value.1/6x + 5/6*(-1) = 0      1/6x = 5/6                   x= 5  So the amount should be $5 to make the game fair.