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?
@joe-shmo saidThe number of tails observed is binomially distributed in each case, with a different value for the parameter p:
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?
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
@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
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