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Licemer1 [7]
2 years ago
5

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, on

e 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.) Xi 2 –1 P(xi) What is Miguel’s expected value from playing the game?
Mathematics
1 answer:
creativ13 [48]2 years ago
6 0

Answer:

The answer is "\bold{-\frac{1}{2}}"

Step-by-step explanation:

Miguel's choose the best way to win the 2 dollars by pulling the two chips with the number 1.  In total, there are four balls, so his probability of winning is:  

\to \frac{2}{4} \times \frac{1}{3}= \frac{1}{6}

The chances of losing a dollar add up to that amount, so 5/6.  

Price predicted = \frac{1}{6} \times (2) + \frac{5}{6}  \times (-1)

                          =\frac{1}{6} \times 2 + \frac{5}{6}  \times -1\\\\=\frac{1}{3} - \frac{5}{6}   \\\\=\frac{2-5}{6}\\\\=\frac{-3}{6}\\\\=- \frac{1}{2}\\

The final answer is "-\frac{1}{2}"

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