# another coin flip question [closed]

I have three coins in a bag. The first coin flips heads with probability 50%, the second coin flips heads with probability 60%, and the third coin flips heads with probability 70%. I pull out a coin and flip heads. If I flip this coin again, what is the probability I will get heads.

This is what I have tried: the sum of probability is: 0.5 + 0.6 + 0.7 = 1.8

$\frac{0.5*0.5}{1.8} + \frac{0.6*0.6}{1.8} + \frac{0.7*0.7}{1.8} = 0.611..$

is there a better way to do this ? ex. how do we do this in bayes theorem ?

## closed as off-topic by gebruiker, Arnaldo, Daniel W. Farlow, Namaste, SmylicJun 8 '17 at 16:02

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You got this from here, didn't you? ;)

The answer as given on the page is right, using Bayes' rule: $$P(HH|H)=\frac{P(HH\wedge H)}{P(H)}$$ $$=\frac{P(HH)}{P(H)}$$ $$=\frac{\frac13(.5^2+.6^2+.7^2)}{\frac13(.5+.6+.7)}=\frac{11}{18}$$

• Thank you this is what I'm looking for, especially the bottom part. – szd116 Jun 7 '17 at 19:50

So the probability you pick out this particular coin is $\frac{1}{3}$.

Finally, as you said, the probability is $\frac{1}{3}(0.5^2+0.6^2+0.7^2)=0.3667$

Three paths.

Pick out the first coin with probability 1/3, flip heads 1/2, flip heads 1/2 probability.

Pick out the second coin with probability 1/3, flip heads 3/5, flip heads 3/5 probability.

Pick out the third coin with probability 1/3, flip heads with 7/10, flip heads 7/10 probability.

So that's where $\frac{1}{3}(0.5^2+0.6^2+0.7^2)$ came from, if you were wondering.

Dividing by 1.8 does not make sense because how often they flip heads should not influence the likelihood of you pulling them out of the bag.

Update: I completely misread the question.

If it is given that the first flip was a head, the second flip has nothing to do with the first flip.

The probability of getting a head is thus $\frac{1}{3}(0.5+0.6+0.7)=0.6$.

• also what the heck is bayes theorem – Saketh Malyala Jun 7 '17 at 18:54
• But you are not conditioning on the fact that the first flip is already a head, that information is useful, but in your model, you are treating it as if it is totally irrelevant. – szd116 Jun 7 '17 at 18:59
• ok i updated it – Saketh Malyala Jun 7 '17 at 19:00