SOLUTION: Roll a die once. Then roll it as many times as the outcome from the first roll. Getting the special number "3" on any roll means a win. What is the expected number of wins from thi
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-> SOLUTION: Roll a die once. Then roll it as many times as the outcome from the first roll. Getting the special number "3" on any roll means a win. What is the expected number of wins from thi
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Question 1178261: Roll a die once. Then roll it as many times as the outcome from the first roll. Getting the special number "3" on any roll means a win. What is the expected number of wins from this experiment? Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! Let X = r.v. equal to the value of the die on the 1st throw.
Then for X = 1, 2, 3, 4, 5, 6, it can be shown combinatorially that +...+.
The probability of a '3' turning up on any throw is 1/6. Hence the expectation for the number of wins in this game is
