If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: High school statistics>Unit 6

Lesson 3: Conditional probability

# Conditional probability with Bayes' Theorem

Conditional probability visualized using trees. Created by Brit Cruise.

## Want to join the conversation?

• This is awesome! Your videos will really help confused students over at Udacity's stats class--this is one of the first things they learn. :) Thank you for this!
• Just got stuck on udacities 'Bayes Rule' chapter and decided to look at KA! :)
• At , Brit says that the probability of Bob picking out the fair coin was 1/3. But since he picked 1 out of 2 coins, fair and unfair, at random, wouldn't the probability just be 1/2?
• Yes, picking one out of the two coins at random would result in a 1/2 probability of having picked the fair coin. However, the question was, what is the probability of having picked the fair coin, GIVEN THAT the coin came up heads. As the title "Conditional Probability" suggests, the probability of having picked the fair coin is dependant on the evidence we have (it came up heads)

Consider the opposite scenario - the coin comes up tails when flipped. Before tossing it, you would be correct in saying there's a 1/2 chance you're holding the fair coin. But given the new evidence we have from the flip - it came up tails - you could be 100% certain of having the fair coin, since the double sided coin only has heads. The probability of the event (you picked the fair coin) is dependant on the evidence (the coin came up tails).
• I just didn't understand the "rules" of the "game". He picks a random coin for each flip OR he picks a random coin only at the beginning, continuing with that one to the rest of the problem?
• random coin for every flip
• If f I have an unfair coin, with a possibility of 2/3 to land on T and 1/3 to land on H, and I flip it 3 times, what are the chances of me to result with 2 Tails and 1 Head?
The order isn't important. Heads can be the first coin, the second or the third.
• Let's pretend it's fair, first.
The long way, you can just list them. Three flips can be
HHH
HHT, HTH, THH
HTT, THT, TTH
TTT
There are 8 possibilities, and there are 3 ways to make 2T 1H, so it's 3/8.

To use math, you could do this: (p represents H, q represents T, and both are 1/2)
p^3 + 3p^2q + 3pq^2 + q^3 (notice how this matches with the possibilities listed)
This is the binomal expansion of (p+q)^3, 3 being the number of tosses you have. If you want a different number of tosses, you just change the exponent for (p+q)^n.

This will always add to one, because that represents 100% of the possibilities. Each section represents the odds of a particular possibility. Since you want 2 tails and 1 head, you choose the one that includes pq^2.
3(1/2)(1/2)^2 = .375, which is equal to 3/8, same as before

Now that I've demonstrated that the equation works, you can substitute any probability in for p and q, as long as they add up to 1. You want p=1/3 and q=2/3, which gives us
3pq^2 = 3(1/3)(2/3)^2 = .4444 or 4/9. So the chances of getting 2 tails and 1 heads in three flips is 4/9, or about 44%.
• This was not worded well. I should have said what is the probability that out of the 2 coins the fair coin was flipped. It said what is the probability he picked the fair coin
• At , he changes the amount of leaves for each of the three coins to six leaves each. Why is this necessary? I thought you can find the probability from the original probabilities (1/2 and 2/3, in this case)? Thanks
• He just wants to show it more visually for those who have problems understanding
• How come we cut the "tails" branch? Why not leave it and use it to calculate the probability?
• Another guess (similar to Sky's):
You can cut the "tails" branches because the outcome of the toss has already been decided, and you know in advance tails is not possible. This tells you that you should not consider "tails" in your list of possible outcomes when you're calculating the probability (probability being the ratio of all favorable outcomes to all possible outcomes). In 'regular' probability, you don't know any of the outcomes in advance, so all outcomes are need to be counted as possible when you calculate your ratio.
• P(biased|head) = P(head|biased)P(biased)/P(head) = (2/3)*(1/3)/(1/2) = 4/9. But the tree calculates it to be 4/10. Where did I go wrong ?