Register to reply 
Conditional Probability on type of coin 
Share this thread: 
#1
Oct1812, 10:21 AM

PF Gold
P: 2,308

1. The problem statement, all variables and given/known data
Suppose we have 10 coins such that if the ith coin is flipped, heads will appear with probability i/10, i = 1,2...10. When one of the coins is randomly selected and flipped, what is the conditional probability that it was the coin? 2. Relevant equations Bayes's Formula 3. The attempt at a solution First of all, I can't make any sense out of how choosing a different coin will give a different probability of showing a head? In solving the problem, I used Bayes's relation; [tex] P(\text{5th coin}  \text{heads}) = \frac{P(\text{heads  5th coin})P(\text{5th coin})}{P(\text{heads 5th coin})P(\text{5th coin}) + P(\text{heads  not 5th coin})P(\text{not 5th coin})} [/tex] where [tex] P(\text{heads 5th coin}) = \frac{5}{10}, P(\text{5th coin}) = \frac{1}{10}, P(\text{heads  not 5th coin}) = \frac{P(\text{heads and not 5th coin})}{P(\text{not 5th coin})} = \frac{\frac{9}{10}\frac{1}{2}}{\frac{9}{10}} = \frac{1}{2}, P(\text{not 5th coin}) = 1\frac{1}{10}. [/tex] Putting this together gives the wrong answer. Any ideas? 


#2
Oct1812, 10:43 AM

Sci Advisor
HW Helper
Thanks
P: 5,083

P{coin iH} = P{coin i & H}/P{H}. What is the numerator equal to? How do you find the denominator? RGV 


#3
Oct1812, 10:58 AM

PF Gold
P: 2,308

Oh sorry, how careless of me. It should read 'When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the 5th coin'



#4
Oct1812, 11:23 AM

Sci Advisor
HW Helper
Thanks
P: 5,083

Conditional Probability on type of coin
RGV 


#5
Oct1812, 11:49 AM

PF Gold
P: 2,308




#6
Oct1812, 12:00 PM

Sci Advisor
HW Helper
Thanks
P: 5,083

RGV 


#7
Oct1812, 12:00 PM

PF Gold
P: 2,308

I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said [itex] \sum_{i=1}^{5} P(headcoin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) + (6/10)(1/10) + (7/10)(1/10) + (8/10)(1/10) + (9/10)(1/10) + (1/10) [/itex] I now get the right result  thanks for your help!
EDIT : let i go form 1 to 10. 


#8
Oct1812, 12:07 PM

Sci Advisor
HW Helper
Thanks
P: 5,083

RGV 


#9
Oct1812, 02:08 PM

Sci Advisor
HW Helper
Thanks
P: 5,083

RGV 


Register to reply 
Related Discussions  
Conditional probability  Probability of spotting a downed airplane (really basic)  Set Theory, Logic, Probability, Statistics  3  
Probability question; Conditional probability and poisson distribution  Precalculus Mathematics Homework  1  
[probability theory] simple question about conditional probability  Precalculus Mathematics Homework  1  
HELP geometric probability: area of a square and conditional probability  Precalculus Mathematics Homework  4  
(CONDITIONAL PROBABILITY) Which probability would you expect to be greater?  Precalculus Mathematics Homework  1 