A book has n typos. Two proofreaders, Prue and Frida, independently read the book. Prue catches

A book has n typos. Two proofreaders, Prue and Frida,...

Key Concepts

Binomial Distribution

This concept describes the probability distribution of the number of successes in a fixed number of independent Bernoulli (success/failure) trials. In the context of the question, the number of typos caught by each proofreader (Prue or Frida) follows a binomial distribution since each typo is independently caught with a certain fixed probability.

Independence of Events

The problem relies on the fact that each typo is checked independently by the two proofreaders and that the catching events for Prue and Frida do not affect one another. This independence allows us to model the outcomes for each typo separately and combine probabilities accordingly.

Complement Rule

This principle is used to calculate the probability of an event by subtracting the probability of its complement from one. Specifically, to find the probability that at least one proofreader catches a typo, we calculate one minus the probability that both miss it, which is a direct application of the complement rule.

Conditional Distribution

This involves determining the probability distribution of a random variable contingent on some known conditions or events. In part (b) of the problem, the conditional distribution of the number of typos caught by one proofreader is derived given the total number caught by both, illustrating how information about one event can update beliefs about another.

Distribution of the Sum of Independent Events

When dealing with independent events, the distribution of the number of successes obtained by at least one observer can be determined by combining their independent probabilities. In this case, the overall probability that a typo is caught is derived from the complement of the joint miss probability, leading to a specific binomial distribution for the total number caught by at least one proofreader.

Transcript

00:01 A professional proofrator has a 98 % chance of finding an error in a piece of work other than misspellings, double words, similar errors that are machine detected.

00:20 This particular work has four errors, and we want to find the probability that the proofrator will miss at least one of them.

00:32 So this is going to be a binomial probability because there's only two possible outcomes.

00:39 Find the error or don't find the error.

00:43 So we are going to have to define our probability of success.

00:48 And in part a, we want the probability that the proof reader misses at least one.

00:55 So we're going to define p as the probability of missing an error instead of finding an error.

01:09 So if there is a 98 % chance of finding an error, then the probability, probability of missing the error would be a 2 % chance, or 0 .02.

01:21 Since there are four errors, we have a fixed number of trials, so n is going to be four.

01:28 So if we look at the probability distribution, we might miss no errors, we might miss one error, we might miss two errors, we might miss three errors, or we might miss four errors.

01:44 And we have to determine the probability of each of those.

01:49 So since it is a binomial situation, the probability can be found by using ncx times p to the x times q to the n minus x power.

02:03 And if p is the probability of success, which we're defining as missing the error, then q is the probability of failure, and they must add up to 1, so q is going to be 0 .98.

02:20 And if we use that formula, with n being 4, and x continually changing from 0 to 1 to 2 to 3 to 4, and we use 0 .02 as our p value, and we use 0 .98 as our q value, then we will get the following probabilities.

02:46 0 .9223, 36816 of missing no errors.

02:56 0 .075526 is the probability of missing one error.

03:05 0 .002 -3496 is the probability of missing two errors.

03:16 0 .000 -3 -136 is the probability of missing three errors, and 0 -0 -0 -0 -0 -1 -6 is the probability of missing three errors, and point 0 -0 -0 -1 -6 is the probability of missing all four errors.

03:36 So in part a, we are trying to determine the probability that we will miss at least one.

03:44 So that would be that x is greater than or equal to one.

03:48 So we can do one of two things.

03:51 We can either add up the probability of missing one error plus the probability of missing two errors, plus the probability of missing three errors, plus the probability of missing four errors, which would be these all totaled together.

04:09 Or the other option we can use is because in a probability distribution, if you sum up the p of x column, you always need to get a value of one.

04:21 So if we take one, take away missing no errors, we will find that same value.

04:29 So we could do one minus the probability of missing no errors.

04:33 And i find that to be a little easier.

04:36 So if i do 1 minus 0 .922 -36816, we will get a probability of 0 .0776 3184.

04:53 And we usually round our probability.

04:56 So i'm going to go to four decimal places and say 0776.

05:02 For part b, we want to show that two such proofreaders, so now we're going to have two different proof readers working independently of each other, have a 99 .96 % chance of missing or detecting an error.

05:30 So let's draw a tree diagram for this.

05:33 The first proofreader might find the error or might not make me.

05:47 Miss the error.

05:50 And that first proofreader has a 98 % finding probability and a 0 .02 missing probability.

06:01 Now when it comes time to do the second proof reader after the first one already looked, which is now going to lead us into that tree diagram, that second proof reader might find it or miss it.

06:19 And if the first one missed, the second one might find it or miss it.

06:24 Keeping in mind that the probability of finding an error is 0 .98 and the probability of missing is 0 .02...

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