You have these 10 observations of some event X:
1.03
1.022
1.022
1.01
1.026
1.02
0.95
1.005
1.05
1.023
Here is the sorted list from least to greatest:
0.95
1.005
1.01
1.02
1.022
1.022
1.023
1.026
1.03
1.05
Based on this data alone:
1) It makes sense to say P(X <= 0.95) = 1/10 correct? We might say there is a 1/10 chance that X <= 0.95.
2) It makes sense to say P(X >= 1.01) = 8/10 correct? We might say there is an 8/10 chance that X >= 1.01.
3) My real question: What is the chance that we will see an observation of X >= 1.01 before we see an observation of X <= 0.95? Can this be calculated with this information?
My guess at an answer to #3:
Since P(X >= 1.01) = 8/10 is eight times P(X <= 0.95) = 1/10, can we say that overall the probability that we see X >= 1.01 before we see X <= 0.95 is 8/9? As in:
Event I: 8/9 chance that X >= 1.01 before X <= 0.95
Event II: 1/9 chance that X <= 0.95 before X >= 1.01
Total Probability: 8/9+1/9 = 1, which makes sense since Event I and Event II together are collectively exhaustive, and 8/9 is 8 times 1/9.
Is this correct for #3? If not, or even if so, is there some standard (“named”) statistical procedure/distribution that deals with answering the question posed in #3?
Thanks for your thoughts.
I began to think of the binomial distribution with the common heads or tails coin flipping experiment example...I believe my question in #3 is related to that, and in fact I think the generalization is what applies, the "multinomial" distribution...any thoughts from any learned stat tutors?
waste of waste
To answer your questions, let's go step by step:
1) P(X <= 0.95) = 1/10: Yes, that is correct. Since there is only one observation that is less than or equal to 0.95 out of the total of 10 observations, the probability is indeed 1/10.
2) P(X >= 1.01) = 8/10: No, that is not correct. In this case, there are four observations that are greater than or equal to 1.01 out of the total of 10 observations. Therefore, the probability should be 4/10 or simplifying it, 2/5.
Moving on to your real question, let's calculate the probability of seeing an observation of X >= 1.01 before seeing an observation of X <= 0.95 using the given data:
To do this, let's count the number of observations that fall into each category:
- Number of observations that are greater than or equal to 1.01: 8
- Number of observations that are less than or equal to 0.95: 1
To find the probability, we divide the number of observations that fall into the first category by the total number of observations:
P(X >= 1.01 before X <= 0.95) = 8 / (8 + 1) = 8/9
So your guess is correct! The probability of seeing an observation of X >= 1.01 before seeing an observation of X <= 0.95 is indeed 8/9.
As for the standard statistical procedure or distribution that deals with this specific question, there isn't a named distribution or procedure for this scenario. However, the concept of conditional probability is being used here, where we calculate the probability of one event happening given that another event has already occurred.
I hope this clears up your questions. Let me know if you need further assistance!