# Expected value with condition

1. May 26, 2013

### ParisSpart

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...

2. May 26, 2013

### danago

$$P(X = k) = (1-p)^{k-1}p$$

You can easily find the value of $P(X \ge 6)$ by realising that $P(X \ge 6)=1-P(X<6)$, 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 $k<6$? What about when $k \ge 6$?

3. May 26, 2013

### I like Serena

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.

4. May 26, 2013

### ParisSpart

i found the P(X>=6) but i cant 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...

5. May 26, 2013

### Ray Vickson

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.