Geometric Distribution Probability problem

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SUMMARY

The discussion focuses on calculating the probability of rolling a fair die until a three or four appears, specifically determining the probability that the number of rolls, denoted as Z, is greater than or equal to 3. The probability of rolling a three (p(A)) or a four (p(B)) is established as 1/6 each, leading to the combined probability of p(A U B) = 1/3. The solution utilizes the geometric distribution formula p(Z >= k) = (1 - p)^(k - 1), resulting in a final probability of 4/9 for Z being greater than or equal to 3.

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  • Understanding of geometric distribution
  • Basic probability concepts
  • Familiarity with fair die mechanics
  • Knowledge of probability notation (e.g., p(A), p(B))
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  • Learn how to calculate probabilities using combinatorial methods
  • Explore advanced probability concepts such as conditional probability
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Hiche
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Homework Statement



We roll a fair die until we get a three or a four. Z denotes the number of rolls needed. What is the probability that Z >= 3? (replacement assumed)

Homework Equations



Geometric distribution seems logical here?

The Attempt at a Solution



Let p(A) = p(getting a three) = 1/6 and p(B) = p(getting a four) = 1/6. We want p(A U B) = 1/3 (a three OR a four). Correct?

Now, we know p(Z >= k) = (1 - p)^(k - 1). Is that enough for this question? I got 4/9.
 
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Hi Hiche! :smile:

Yes, 4/9.

Probably best to specifically say that Z ≥ 3 means that there is no 3 or 4 in the first two rolls, the probability of which is P(not A or B) squared. :wink:
 

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