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13

The answers provided here so far give lots of good tips but I think they're not addressing a key part of the question, which is "why do we need to count two events (50,52) and (52,50), instead of one event (50,52)?" The answer is that you can do it either way. The textbook way is counting P(X=50 \cap Y=52) + P(X=52 \cap Y=50) + P(X=51 \cap Y=51) = P(X=...

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This is a very good question. The issue comes up frequently. I explain this using a toy model: throw two regular six-sided die. What is the probability that the sum is 3? With some physical modeling, you can become convinced that the answer is 2/36=5.5%, This corresponds to the two possibilities 1+2 and 2+1. But why are these two possibilities distinct? We ...

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Q1. A real-world story that I've heard many times over when Bayes' Theorem is explained concerns medical lab tests where one of the events in your table would be "actually has the disease the test is designed to detect" while another is "experiences a positive test result when the lab test is administered"; however, these scenarios tend to be confusing for ...

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Emphasize to the student that Every probability has an associated experiment. Every experiment has an associated sample space. Every event is a subset of the sample space. In this case, the associated experiment is "randomly take one ball from a box, note the color, then return the ball to the box. Repeat." The associated sample space is $\{P_1P_1,... 3 I don't think I can improve on juod's excellent answer, but there are a few points worth elaborating. First, physical objects are always distinguishable. (I am here ignoring the phenomenon of identical particles in quantum mechanics, the issue there being that we really have to rethink the concept of physical object when talking about the electrons in a ... 3 The original question is: Four people are randomly chosen from a place. Assuming the birthdays of people are equally likely to occur in any month, find the probability that 4 selected students are not born in the same months. I understand that the intended event includes cases such as [Jan, Feb, Mar, Apr] and [Jan, Feb, Feb, Feb] and excludes cases such ... 2 The succinct answer is that there are two ways to choose$50$and$52$- if two boxes are chosen one could have$50$and other$52$or vice-versa - but this will be more apparent to a student if you write out the space of all$25$possible outcomes as a list of ordered pairs. "Indistinguishable" does not mean "identical", just as "isomorphic" does not mean "... 2 Per Wiki: https://en.wikipedia.org/wiki/Markov_chain#Gambling Suppose that you start with \$10, and you wager \$1 on an unending, fair, coin toss indefinitely, or until you lose all of your money. If$X_n$represents the number of dollars you have after$n$tosses, with$X_0 = 10$, then the sequence$\{X_n: n\in \mathbb{N}\}$is a Markov process. If I ... 2 I don't know from what principles you're teaching probability, but the answer I would give is the following: When calculating probabilities, it is important to remember the basic principles. We have a set of possible outcomes$S$and an event$E \subseteq S$. The probability that an outcome in our event is chosen is precisely$\frac{E}{S}\$, *given that ...

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I think there are a lot of good answers here, but one point of view that might convince some students is to use a simulation - not of the toy problems e.g. with dice suggested, but with this one. Apologies in advance for my terrible R code - live version - here I assume a uniform discrete distribution on 48 to 52. counter = 0 for (i in 1:100000) { ...

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You didn't disclose the level you teach. When I first ran into this issue (i.e. the need to explain this), I went to the example... You have 2 coins. There are 3 possible outcomes, 2 heads, One Head One Tail, 2 Tails. Are they each equally likely, each 1 in 3? I then walk them through the physical experiment. Even though we flip two coins at once (as ...

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Challenge him with the following: let‘s think it through with a soccer match. In how many ways can the match end, such that the sum of the goals is 3? According to his logic only two possible ways (1, 2) or (0, 3) - I‘m pretty sure the fans won‘t agree. Same goes for the colored dice - how many ways can they sum up to 4? Thus what’s the probability for ...

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You could explain that the first box always goes to Alice, while the second goes to Bob. Then it becomes obvious that Alice getting 50 and Bob getting 52 is different then Alice getting 52 and Bob getting 50, so much so that one of them could die if they get the wrong amount. On the other hand, Alice and Bob can both get 51 in only one possible way, and ...

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Its easy for even "seasoned" mathematicians to make mistakes in such straightforward probability questions. With that in mind I suggest the following - Insist on solving the problems systematically (especially for beginners). This means - writing out the sample space (if not in its entirety, at least the pattern), rereading what the event is, choosing the ...

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Judea Pearl & Dana Mackenzie, in their new book The Book of Why (p.190ff), explain the paradox in a way I hadn't seen before. Pearl imagines changing the rules to "Let's Fake a Deal," where "Monty opens one of the two doors you didn't choose, but his choice is completely at random." Of course he could open the door containing the car/prize, ruining ...

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Please warn me if my rewrite of this /r/eli5 comment doesn't answer your bolded question. It helps to get rid of the standard trappings of the problem. The problem beneath is mathematically the same, but makes it clearer why you ought switch. You are a superhero patrolling a crowded train station. A male stranger comes up to you, and asks you to pick a ...

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