Question

student guide (continued)

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.)

b. what is miguels expected value from playing the game?

Explanation:

Step1: Calculate the probability of winning ($X = 2$)

There are \(C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4\times3}{2\times1}=6\) ways to choose 2 chips out of 4.
The only way to win (both chips have the same number) is when we choose the two chips with number 1. There is \(C(2,2) = 1\) way to choose the two chips with number 1.
So \(P(X = 2)=\frac{C(2,2)}{C(4,2)}=\frac{1}{6}\)

Step2: Calculate the probability of losing ($X=-1$)

Using the complement rule \(P(X=-1)=1 - P(X = 2)\)
Since \(P(X = 2)=\frac{1}{6}\), then \(P(X=-1)=1-\frac{1}{6}=\frac{5}{6}\)

Step3: Calculate the expected value \(E(X)\)

The formula for the expected value of a discrete random variable is \(E(X)=\sum_{i}x_{i}P(x_{i})\)
Here \(x_1 = 2\), \(P(x_1)=\frac{1}{6}\), \(x_2=-1\), \(P(x_2)=\frac{5}{6}\)
\(E(X)=(2)\times\frac{1}{6}+(-1)\times\frac{5}{6}=\frac{2 - 5}{6}=-\frac{3}{6}=-\frac{1}{2}\)

Answer:

a. \(P(X = 2)=\frac{1}{6}\), \(P(X=-1)=\frac{5}{6}\)
b. Miguel's expected value from playing the game is \(-\frac{1}{2}\) (or \(- 0.5\))