Question 1197263
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The tutor @ikleyn has a great solution.
My approach is the same idea, just written slightly different. 
There's also an alternative visual way to see how this all works.


H = heads
T = tails


Let's define these four events
A = coin is C1
B = coin is C2
C = coin is C3
D = result is HH


We're asked to find P(A given D)


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First let's take a slight detour


P(A) = P(B) = P(C) = 1/3
assuming each coin is equally likely to be selected
note that
P(A)+P(B)+P(C) = 1


Then calculate the conditional probability of getting HH for each coin
P(D given A) = P(H given A)*P(H given A) = (1/3)*(1/3) = 1/9
P(D given B) = P(H given B)*P(H given B) = (2/3)*(2/3) = 4/9
P(D given C) = P(H given C)*P(H given C) = 1*1 = 1


Those 6 items will then help us determine P(D)


Use the Law of Total Probability to find the following
P(D) = P(D and A)+P(D and B)+P(D and C)
P(D) = P(D given A)*P(A)+P(D given B)*P(B)+P(D given B)*P(B)
P(D) = (1/9)*(1/3)+(4/9)*(1/3)+(1)*(1/3)
P(D) = 1/27 + 4/27 + 1/3
P(D) = 1/27 + 4/27 + 9/27
P(D) = (1+4+9)/27
P(D) = 14/27
Thing to notice: In the diagram below, exactly 14 rectangles are shaded yellow out of 9*3 = 27 total.

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Now we can calculate what we're after.
P(A given D) = P(A and D)/P(D)
P(A given D) = P(D given A)*P(A)/P(D)
P(A given D) = (1/9)*(1/3)/(14/27)
P(A given D) = (1/9)*(1/3)*(27/14)
P(A given D) = (1*1*27)/(9*3*14)
P(A given D) = (1*1*27)/(27*14)
P(A given D) = <font color=red>1/14
which is the final answer.</font>


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Here's a visual way to think about it
<img src = "https://i.imgur.com/xH5vh2H.png">
C1,C2,C3 are equally likely. They represent the three columns.
The numbers inside the yellow boxes help us keep track of each column's values.
Each yellow rectangle represents the outcome HH, while a white nonshaded rectangle is some other outcome (HT, TH or TT).


Column C1 has 1 yellow rectangle shaded out of 9. This is to represent this coin's probability of getting two heads in a row.


Column C2 has 4 yellow rectangles shaded out of 9, showing that 4/9 is the probability of getting HH for coin C2.


Column C3 has all 9 rectangles shaded since we're guaranteed to get two heads for this coin.


There are 1+4+9 = 14 rectangles shaded yellow.


Now imagine we know 100% that we got two heads in a row. That must mean we know we landed somewhere on a yellow rectangle, or we know for certain we have pulled a yellow rectangle out of the bag. 


C1 only chips in one such shaded rectangle, so that's another way to see how <font color=red>the answer is 1/14</font>


If we wanted to know P(B given D), then the answer would be 4/14 since C2 chips in 4 yellow rectangles out of the 14 total.


Lastly P(C given D) = 9/14 for similar reasoning.


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<font color=red>Answer: 1/14</font>
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