What Is the Expected Value E(X|X>=6) for a Geometric Distribution with p=0.20?

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Homework Help Overview

The discussion revolves around finding the expected value E(X|X>=6) for a geometric distribution with a probability parameter p=0.20. Participants are exploring the implications of the condition X>=6 and how it affects the expected value calculation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for expected value and raise questions about how to compute probabilities P(X=k|X>=6) and P(X>=6). There is an exploration of the meaning of the condition X>=6 and its implications for the expected number of additional throws required.

Discussion Status

Some participants have successfully calculated P(X>=6) but are struggling with P(X=k AND X>=6). There is an ongoing examination of the events corresponding to different values of k in relation to the condition X>=6, indicating a productive exploration of the problem.

Contextual Notes

Participants are working under the constraints of a geometric distribution and are considering the implications of the condition X>=6 on their calculations. There is an emphasis on understanding the events associated with different values of k before delving into probability calculations.

ParisSpart
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the random value X takes values 1,2,3... and has the X has geometric distribution with p=0.20 (This means that X can be interpreted as the time the first crown to repeated throws a coin coin lands heads with probability p.) what is the expected value E(X/X>=6)=?

i use this type : E(X/X>=6)=sum(k*P(X=k/X>=6))
but how i can estimate this> how i will find P(X=k and X>=6)=? and P(X>=6)=? where k=1,2,3...
 
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<br /> P(X = k) = (1-p)^{k-1}p<br />

You can easily find the value of P(X \ge 6) by realising that P(X \ge 6)=1-P(X&lt;6), which is computationally simple to evaluate (or use the equation for the cumulative distribution function, which you can find with a quick Google search).

P(X=k \cap X \ge 6) is quite simple if you consider particular values of k. What is its value if k&lt;6? What about when k \ge 6?
 
What does the condition X>=6 mean in words?
What can you say about expectation of the number of additional throws required?

If you understand it, you can see the answer immediately.
 
i found the P(X>=6) but i can't find the P(X=k AND x>=6) i sum all them :sum from k=6 to inf of p*(1-p)^(k-1) bt its not correct...
 
ParisSpart said:
i found the P(X>=6) but i can't find the P(X=k AND x>=6) i sum all them :sum from k=6 to inf of p*(1-p)^(k-1) bt its not correct...

For k < 6, what is the event {X = k and X >= 6}? (Never mind about probabilities for now; just answer my question about the event. For k = 6, what is the event {X = k and X >= 6}? For k > 6, what is the event {X = k and X >= 6}?

As I said, answer these questions first, then worry afterwards about probabilities.
 

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