You are playing Super Mario Bros together with 2 of your friends. You got to level 4, where you encounter your nemesis Bowser. Bowser is very strong, and he is defeated only 41% of the times. Each of you will play level 4 one time. Suppose that, after your friends are gone, you decide to play level 4 until you beat Bowser. Let Y be the number of times you play level What is the distribution of Y? What is the probability that you play less than 3 times? What is the expected number of times that you play?
You are playing Super Mario Bros together with 2 of your friends. You got to level 4, where you encounter your nemesis Bowser. Bowser is very strong, and he is defeated only 41% of the times. Each of you will play level 4 one time.
Suppose that, after your friends are gone, you decide to play level 4 until you beat Bowser. Let Y be the number of times you play level
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What is the distribution of Y?
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What is the probability that you play less than 3 times?
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What is the expected number of times that you play? - for this answer I got 2.439 because I did E(y)=1/p = 1/0.41 = 2.439
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Let defeating the nemesis be termed as a success, the probability of success, p = 0.41
and the probability of failure, q = 1-p = 0.59
If Y be the random variables denoting the number of times one should play until first success, then the probability of success in y trials is given by
qy-1p
Part a: The distribution of Y is Geometric distribution where
P(Y = y) = qy-1p, y = 1, 2, 3, ...
Part b: The required probability is
P(Y < 3) = P(Y = 1) + P(Y = 2)
= q1-1p + q2-1p
= 0.41 + 0.590.41
= 0.6519
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