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Hello again Dr. Oliver, I wanted to post the question again as it was closed with the assumption that it was a homework problem. I am a professional working in industry and would like for someone to assess my work and provide some guidance if I am headed in the right direction with my analysis of this question. There are variables in the "big picture" that I did not explain for the sake of brevity. This is a question I was asking myself after looking at our data set and wanted to summarize the distribution using a simple number. I figure I'll also mention that we are not dealing with Postal services or letters, rather just a cover-up for data packets over the network ;).

Question:

A Postal service received 900 letters, but was only able to deliver 700 of them. Of these 700 delivered, the average arrival time was 6 days. Only 500 (of the delivered letters) arrived within 5 days. What is the probability that, a letter will be delivered successfully, AND within 5 days?

P(A) = P(Successfully deliver a letter) = 7/9

P(B) = P(Successfully deliver a letter within 5 days) = 5/7

P(Successfully deliver a letter AND within 5 days) = P(A) * P(B|A)

P(B|A) = P(B) because this is a subset of the successfully delivered letters, but inclusive to 0-5 days?

so (7/9) * (5/7) = 5/9 = .55 = 55.6% ?

Please advise. Thank you.

Alright, this is a very simple question but I'm going to have to be especially clear. You have the right answer, but for the wrong reasons. First of all, you are really badly defining your events. You are using three events here:

Successfully deliver a letter

Successfully deliver a letter within 5 days

Successfully deliver a letter AND within 5 days

How is the third event different from the second???

Define your events clearly and precisely:

Event A = successful delivery

Event B = successful delivery within 5 days

[third event not listed due to redundancy]

Another very important distinction: P(B|A) is not P(B), not now, not ever (unless A is trivial). The correct formula is:

P(B|A) = P(B and A) / P(A)

If you work in computers, you should be happy with "and" being ∩, the intersection, so:

P(B|A) = P(B∩A) / P(A)

You sort of wrote something like this, but you've got all the terminology mixed up, and you've defined all your events in inconsistent ways. So let's be clear

Now it is true that B ⊆ A, thus B∩A = B. This means:

P(B|A) = P(B) / P(A)

Instead of saying P(B∩A)=P(B), you mistakenly say P(B|A)=P(B). So let's review:

P(B) = 5/9, not 5/7. Of all 900 letters, only 500 arrive within 5 days. P(B) must account for the entire set of letters, so the denominator is 9, not 7.

P(B∩A) = P(B) because out of all letters, 500 are delivered and are delivered within 5 days (because the condition of being in A is redundant if we are already in B).

P(B|A) is the only thing you could be solving for here, so let's say what this means. This means "ASSUMING the letter is delivered, what is the probability it is there within 5 days?" And that is P(B∩A)/P(A). You can read more here about conditional probability.

Now that we are clear, you get:

P(B|A) = P(B)/P(A)

P(B) = 5/9

P(A) = 7/9

P(B|A) = (5/9) / (7/9) = 5/7.

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Comment | Fantastic. Dismissed me at first but provided a very thorough explanation and guided me to the right approach. I am very appreciative of Dr. Oliver's help. |

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

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