In a board game, players take turns spinning a wheel with 4 spaces and values of \$100, \$300, \$400, \$800. The probability of landing on \$100

Question

In a board game, players take turns spinning a wheel with 4 spaces and values of \$100, \$300, \$400, \$800. The probability of landing on \$100 is 4-9. The probability of landing on \$300 is 2/9. The probability of landing on \$400 is 2/9. The probability of landing on \$800 is 1/9. What is the expected value of spinning the wheel once

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3 weeks 2021-11-18T10:35:35+00:00 2 Answers 0 views 0

\$1,444.45

Step-by-step explanation:

EV= 100(0.44444444444)+300(0.22222222222)+400(0.22222222222)+800(0.11111111111)

EV= \$288.89

288.89*5= \$1,444.45

2. The expected value  defined as E(X) = Σ[ X × p(X)] of spinning the wheel once is \$288.89

Creating the distribution table :

X ___ 100 __ 300 ___ 400 __ 800

P(X)_ 4/9 __ 2/9 ____ 2/9 ___ 1/9

E(X) = [(100 × 4/9) + (300 × 2/9) + (400 × 2/9) + (800 × 1/9)]

E(X) = 400/9 + 600/9 + 800/9 + 800/9

E(X) = 2600/9

E(X) = 288.89

Hence, the Expected value of spinning the wheel once is \$288.89