A Tail of Two Coins

There are two coins in a bag. The one coin is unbiased, whereas the other coin is biased with the probability of obtaining a tail of 1/3.
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Your friend chooses one of the coins at random and tosses it 5 times.

You ask your friend, “Did you observe at least three tails?”

Your friend replies, “Yes.”

What is the probability that the biased coin was chosen?

@joe-shmo said
There are two coins in a bag. The one coin is unbiased, whereas the other coin is biased with the probability of obtaining a tail of 1/3.
​
Your friend chooses one of the coins at random and tosses it 5 times.

You ask your friend, “Did you observe at least three tails?”

Your friend replies, “Yes.”

What is the probability that the biased coin was chosen?

@ashiitaka said
The number of tails observed is binomially distributed in each case, with a different value for the parameter p:

X(unbiased) ~ B(5, 1/2)
X(biased) ~ B(5, 1/3)

You can either use the appropriate mass function and add the probabilities separately or the cumulative distribution function to find the probability of X >= 3 for both, which is 1/2 and 17/81 respectively.

P(biased) = (17/81)/((1/2) + (17/81)) = 34/115

Reveal Hidden Content

I wrote up a solution that examines what went on behind the scene in Ashiitaka's solution...but robo mod deemed it inappropriate?!?

Round 2:

let 𝑈 be the event of selecting the unbiased coin from the bag
Let 𝐵 be the event of selecting the biased coin from the bag

P( 𝑈 ) = P( 𝐵 ) = 1/2

Let the probability of getting at least three tails given 𝑈 be:

 P( ≥ 𝑇𝑇𝑇 | 𝑈 ) = ( 1/2 )⁵ ∙ ( C( 5,3 ) + C( 5,4 ) + C( 5,5 ) )

       = ( 1/2 )⁵ ∙ ( 10 + 5 + 1 )

       = 1/2

Let the probability of getting at least three tails given 𝐵 be:

 P( ≥ 𝑇𝑇𝑇 | 𝐵 ) = ( 1/3 )³ ∙ ( C( 5,3 ) ∙ ( 1/3 )³∙ ( 2/3 )² + C( 5,4 ) ∙ ( 1/3 )⁴ ∙ ( 2/3 ) + C( 5,5 ) ∙ ( 1/3 )⁵ )

       = ( 1/3 )⁵ ∙ ( 40 + 10 + 1 )

       = 17/81

Finally, the probability the biased coin was chosen is given at least three tails observed:

 P( 𝐵 | ≥ 𝑇𝑇𝑇 ) = P( 𝐵 ) ∙ P( ≥ 𝑇𝑇𝑇 | 𝐵 ) / [ P( 𝐵 ) ∙ P( ≥ 𝑇𝑇𝑇 | 𝐵 ) + P( 𝑈 ) ∙ P( ≥ 𝑇𝑇𝑇 | 𝑈 ) ]

       = ( 1/2 ) ∙ ( 17 / 81 ) / [ ( 1/2 ) ∙ ( 17 / 81 ) + ( 1/2 ) ∙ ( 1/2 ) ]

       = 34/115