On the first roll, you have a 1/3 probability of rolling a 1 or 2 and thus winning $200.
There's a 2/3 probability of not rolling a 1 or 2 on the first roll. On the second roll, there is again a 1/3 probability of rolling a 1 or 2, and 2/3 probability otherwise, so that there is a 1/3*2/3 = 2/9 probability of getting a 1 or 2 and thus winning $100, and 2/3*2/3 = 4/9 probability of losing.
(a) Let be a random variable representing the winnings from playing the game. It has the probability mass function
(b) Compute the expected value of :
Answer:
22
Step-by-step explanation:
sup b
Answer:
2,984,736
Step-by-step explanation:
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