Thu Jan 17, 2013 11:30 pm by tartle 


A jar contains 3 coins. 2 are heads and tails. 1 is heads and heads.
You pick out at coin and toss it 3 times. You get three heads.
Question ; What is the probability of getting a head on the fourth toss ? 




Sun Sep 08, 2013 9:18 pm by L 


3/8 using a probability tree.
Could not be bothered to try the binomial theorem as I have no tables. 




Tue Dec 24, 2013 6:49 pm by Twerp 


Me thinks it be 2/3. 




Thu Jul 14, 2016 6:35 am by Dzallen 


Probability of getting 4 heads from the start:
P(A)=2*(1/3)*(1/16)+1/3=9/24
Probability of getting 3 heads from the start:
P(B)=2*(1/3)*(1/8)+1/3=10/24
Probability of getting 4 heads given that 3 heads flipped already:
P(A given B)=P(A)/P(B)=9/10
The probability of another head is 9/10
Last edited by Dzallen on Mon Oct 31, 2016 9:08 am; edited 1 time in total 




Thu Jul 14, 2016 2:30 pm by bathfilms 


No need for probability trees, or even looking at the history of the first 3 throws!
The question only asks about the fourth toss.
There is
1/3 probability of coin A (with p(heads) = 1/2) being used
OR
1/3 probability of coin B (with p(heads) = 1/2) being used
OR
1/3 probability of coin C (with p(heads) = 1) being used
P(A OR B OR C) = P(A) + P(B) + P(C)
= (1/3 * 1/2) + (1/3 * 1/2) + (1/3 * 1)
= 2/3 




Thu Jul 14, 2016 2:30 pm by bathfilms 


No need for probability trees, or even looking at the history of the first 3 throws!
The question only asks about the fourth toss.
There is
1/3 probability of coin A (with p(heads) = 1/2) being used
OR
1/3 probability of coin B (with p(heads) = 1/2) being used
OR
1/3 probability of coin C (with p(heads) = 1) being used
P(A OR B OR C) = P(A) + P(B) + P(C)
= (1/3 * 1/2) + (1/3 * 1/2) + (1/3 * 1)
= 2/3 




Fri Jul 15, 2016 1:29 am by Dzallen 


bathfilms, yes you do need to look at the first three throws, because the first 3 heads tell you that it is more likely you are holding a head and head coin rather then a head and tail coin. In other words, the probabilities of having the three coins are not 1/3 each.
For example, if you started with 1 head and head coin, but 2 tail and tail coins, and you throw three heads, the probability of a 4th head is clearly 1, but by your calculation it would be 1/3.
This is a question in conditional probability, the solution is as I posted above. 




Fri Jul 15, 2016 8:25 am by bathfilms 


Dear Dzallen,
Yes, thank you, I see your point of view, which is of course correct.
So, if the 1st three tosses were heads, but you had your eyes closed while sodoing, that would change the probability. It's a bit like Monty Hall's dilemma. 




Mon Oct 31, 2016 2:37 am by Sagitaur 


Surely the answer is 2/3. Label the coins A and B (each has a head & tail) and C for the two headed coin.
The probability of selecting coin A is 1/3. The probability of throwing a head regardless of previous throws is 1/2. The probabilit of a Head if Coin A is 1/3 multiplied by 1/2 = 1/6. The same applies for Coin B.
For coin C the probability is 1/3 to select it and then 1 to throw a head = 1/3. So the probability of a head is 1/6 or 1/6 or 1/3. For 'or' results you add the probabilities ie 1/6+1/6+1/3 =2/3 




Thu Apr 26, 2018 8:48 pm by bedead 


2/3 






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