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?

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

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