Conditional probability (Bayes' Theorem)
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when answering coin-toss and biased-coin questions where a probability tree and Bayes' theorem make the reasoning clear.
Recommended by 2 people across 2 different months, 2016 to 2018.
What it is
A five-minute animated lesson from Art of the Problem that introduces conditional probability and Bayes' theorem. It uses a tree diagram analogy to show how new evidence changes our confidence in earlier events, serving as a primer on Bayesian statistics.
What people said
Recommends Grey Matters · 2017-10 · Mathematics Stack Exchange · on “Tossing coins, one fair, one unfair: how to represent the probability tree?”
You have to remember that the total possibilities include 4 different sides: the fair coin's heads, the fair coin's tails, the unfair coin's head side A, and the unfair coin's head side B. When calculating probabilities, you must, of course, consider total possibilities, and that includes the two possible different heads. Just because you can't distinguish between them, doesn't mean they aren't relevant. Your first tree is the way to go for working out probabilities correctly. Art of the Problem's Bayes Theorem video [this video] explains this exact problem in more detail. For the more general problem of why you should take different sides into account, even when they're the same, read Martin Gar
Recommends Parthiban Rajendran · 2018-03 · Mathematics Stack Exchange · on “What is p(biased coin given heads) in 2 Fair coin, 1 biased coin experiment”
In "Conditional probability with Bayes" [this video] theorem in Khan's academy, in 2nd experiment, where author has 2 fair coins, and 1 biased coin, he tries to calculate probability of biased coin, after first test: Heads. Note : A biased coin is one which gives heads 2/3 times, and tail 1/3 times p(B) = probability of biased coin p(B/H) = probability of biased coin, given its Heads. p(H) = probability of coin being Heads p(H/B) = probability of coin being Heads, given its biased p(B and H) = probability of coin being both biased and heads. Event 1: Pick: p(B) = 1/3 (Since he has 2 fair coins and 1 biased coin) Event 2: Flip: Heads: p(H/B) = 2/3 (Since a biased coin gives heads 2/3 of the time)
Cites ClownInTheMoon · 2016-11 · Mathematics Stack Exchange · on “Combining coin toss probabilities”
This answer is not correct. To answer this question let's first think about the first flip, and the probability that we have selected a fair coin. Now, there are a total of $2*84=168$ equally likely outcomes on the first flip (with the biased coin accounting for two of those outcomes since there are two sides of the biased coin showing $H$)... but we are given evidence that the coin flip is heads. Thus, we can eliminate some of the outcomes on our first flip. For all of the fair coins, of which there are $83$, we can eliminate all of the outcomes in which one of the coins was tails, thus there are $83$ ways we could've gotten to a heads depending on the fair coin we chose. With the biased co