Question

1. 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). a. 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.) Xi 2 –1 P(xi) b. What is Miguel’s expected value from playing the game? c. Based on the expected value in the previous step, how much money should Miguel expect to win or lose each time he plays? d. What value should be assigned to choosing two chips with the number 1 to make the game fair? Explain your answer using a complete sentence and/or an equation.

Answer 1

Answer: hi do you have the answers and work to the rest of this assignment (the student guide one from edge)? if u do can u let me know and i can give u my info to send it to me i would appreciate it so much!

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Given that Miguel is playing a game

The box contains 4 chips, 2 with number 1, and other two differntly numbered as 3 and 5.

OUt of these 4, 2 chips are drawn

P(drawing same number) = 2C2/4C2 =$\frac{1}{6}$

Prob (drawing differnt numbers) = 1-1/6 =$\frac{5}{6}$

Hence prob of winning 2 dollars = $\frac{1}{6}$

Prob of losing 1 dollar = $\frac{5}{6}$

b) Expected value = sum of prob x amount won

= $\frac{1}{6}2+\frac{5}{6}(-1)=-\frac{1}{2}$

c) Miguel can expect to lose 1/2 dollars for every game he plays

d) If it is to be a fair game expected value =0

i.e. let the amount assigned be s

Then $\frac{1}{6}s-\frac{5}{6}=0\\s=5$