Guest Posted December 10, 2010 Report Share Posted December 10, 2010 Hello, my first question on here. We have 3 coins - A, B and C. Coin A has probability of Head (H) = 60 and Tail (T) = 40. If we flip all the coins, what is the probability of geting Heads on coin A. Answer is 3/8 but I get something else. Thanks! Quote Link to comment Share on other sites More sharing options...
0 k-man Posted December 10, 2010 Report Share Posted December 10, 2010 but the probability of coin A showing heads doesn't depend on the other coins. It won't change whether you flip it alone or with other coins. So, the answer to the question as you phrased it is 60%. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 10, 2010 Report Share Posted December 10, 2010 but the probability of coin A showing heads doesn't depend on the other coins. It won't change whether you flip it alone or with other coins. So, the answer to the question as you phrased it is 60%. What you said perfectly makes sense. I have part of the solution which involves, (1/3x0.5)+(1/3x0.5)+(1/3x0.6). Answer is based on Bayes theorem. Final answer is 3/8. Even I find it odd overall but that is what it is. Earlier I thought it would be 1/3 x 0.6 since chance of selecting coin A is 1/3 and that times 0.6 for heads. Maybe I'm just trying to figure out a question that gives me the answer 3/8. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 10, 2010 Report Share Posted December 10, 2010 ok this problem is confusing me Bayes law deals with conditional probability.. P(A|B) = [P(B|A)*P(A)]/P(B) but in the case of this problem as its stated, the only condition is that all of the coins were flipped. If thats the case, wouldnt the probability of coin A flipping heads be independant of the probability that coins B and C flips heads since we dont know (or care) what coins B and C landed on. Like for instance, the probabilty of A flippin heads is the same (p=0.6) regardless of what B and C flip because there is only one event happening and the coin flips are independant. If we were asked what the probabilty was that A flipped heads GIVEN that B and/or C flipped something specific, that would involve Bayes law. I could be wrong on this though if you know for a fact that the answer is 3/8 so i would be interested in knowing the solution to this. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 10, 2010 Report Share Posted December 10, 2010 (edited) What you said perfectly makes sense. I have part of the solution which involves, (1/3x0.5)+(1/3x0.5)+(1/3x0.6). Answer is based on Bayes theorem. Final answer is 3/8. Even I find it odd overall but that is what it is. Earlier I thought it would be 1/3 x 0.6 since chance of selecting coin A is 1/3 and that times 0.6 for heads. Maybe I'm just trying to figure out a question that gives me the answer 3/8. Apologies but I think now I am reaching towards the right question. All coins are flipped and we are interested in a heads on coin A. My solution as below: It's conditional probability - i.e. prob of A getting heads while selecting all 3 coins - so we deal with each coin as well. Further (1/3x0.5)+(1/3x0.5)+(1/3x0.6) = 8/15 (...say I) Probability of only selecting heads coin A with heads on = 0.6/3 (....say II) Divide I by II = (1/5) / (8/15) = 3/8 Do you guys think it's right? Edited December 10, 2010 by digisync Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 10, 2010 Report Share Posted December 10, 2010 What is the probability that, if a randomly selected coin is flipped and turns up heads, that it is coin A, Alex? Quote Link to comment Share on other sites More sharing options...
0 k-man Posted December 10, 2010 Report Share Posted December 10, 2010 What you said perfectly makes sense. I have part of the solution which involves, (1/3x0.5)+(1/3x0.5)+(1/3x0.6). Answer is based on Bayes theorem. Final answer is 3/8. Even I find it odd overall but that is what it is. Earlier I thought it would be 1/3 x 0.6 since chance of selecting coin A is 1/3 and that times 0.6 for heads. Maybe I'm just trying to figure out a question that gives me the answer 3/8. This sounds like homework. There is a forum on this site that helps with homework Homework Help. If this is homework, please post your question there. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 10, 2010 Report Share Posted December 10, 2010 This sounds like homework. There is a forum on this site that helps with homework Homework Help. If this is homework, please post your question there. Isn't a homework question but came across this in a book I was reading (but I'm not a math student lol)...was curious to find out what the solution is. But you think my solution is right? It's the first time I found out about Bayes theorem. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 11, 2010 Report Share Posted December 11, 2010 As the problem was stated, the probability of a heads on coin A is 0.6. Doesn't matter how many other coins there are. Quote Link to comment Share on other sites More sharing options...
Question
Guest
Hello,
my first question on here.
We have 3 coins - A, B and C. Coin A has probability of Head (H) = 60 and Tail (T) = 40. If we flip all the coins, what is the probability of geting Heads on coin A.
Answer is 3/8 but I get something else.
Thanks!
Link to comment
Share on other sites
8 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.