John has 2 coins, one fair and the other unbalanced so that the probability of its coming up heads is 2/3. He picks one of the coins at random, tosses it, and it comes up heads. What is the probability that he picked the unbalanced coin?

To solve this problem, we can use Bayes' theorem. Let's define the events as follows:

A: John picked the fair coin.
B: John picked the unbalanced coin.
H: The coin comes up heads.

We are interested in finding the probability of event B (picking the unbalanced coin) given event H (the coin comes up heads), denoted as P(B|H).

According to Bayes' theorem, we have:

P(B|H) = (P(H|B) * P(B)) / P(H)

P(H|B) represents the probability of obtaining heads given that John picked the unbalanced coin. This is simply 2/3 because the unbalanced coin has a 2/3 chance of coming up heads.

P(B) represents the probability of John picking the unbalanced coin. Since he picks one of the two coins at random, P(B) = 1/2.

P(H) represents the probability of getting heads. This can be calculated using the law of total probability:

P(H) = P(H|B) * P(B) + P(H|A) * P(A)

P(H|B) * P(B) represents the probability of getting heads when picking the unbalanced coin multiplied by the probability of picking the unbalanced coin. This is (2/3) * (1/2) = 1/3.

P(H|A) represents the probability of getting heads when picking the fair coin. This is simply 1/2 because the fair coin is unbiased.

P(A) represents the probability of John picking the fair coin. P(A) = 1/2, just like P(B).

Therefore:

P(H) = (2/3) * (1/2) + (1/2) * (1/2) = 1/3 + 1/4 = 4/12 + 3/12 = 7/12.

Now we can substitute these values back into Bayes' theorem:

P(B|H) = (2/3 * 1/2) / (7/12) = (2/6) / (7/12) = (2/6) * (12/7) = 24/42 = 4/7.

Hence, the probability that John picked the unbalanced coin given that it came up heads is 4/7.

To answer this question, we can use Bayes' theorem.

Let's define the following events:
A: John picked the fair coin
B: John picked the unbalanced coin
H: The coin comes up heads

We are asked to find the probability of event B given that H occurred, denoted as P(B|H).

According to Bayes' theorem, we have the formula:

P(B|H) = (P(H|B) * P(B)) / P(H)

Now let's break down the different probabilities:

1. P(H|B): This is the probability of getting heads given that the unbalanced coin is chosen. We are told that the probability of getting heads with the unbalanced coin is 2/3.

2. P(B): This is the probability of initially picking the unbalanced coin. Since John randomly selects one of the two coins, the probability of selecting the unbalanced coin is 1/2.

3. P(H): This is the probability of getting heads, regardless of which coin is selected. To calculate this, we need to consider two cases:
a) John picks the fair coin and gets heads, denoted as P(H|A) * P(A)
b) John picks the unbalanced coin and gets heads, denoted as P(H|B) * P(B)
We know that the probability of getting heads with the fair coin is 1/2 (since it is a fair coin), and the probability of randomly selecting the fair coin is also 1/2.

Now we can substitute these values into the formula:

P(B|H) = (P(H|B) * P(B)) / (P(H|A) * P(A) + P(H|B) * P(B))

P(B|H) = (2/3 * 1/2) / (1/2 * 1/2 + 2/3 * 1/2)

P(B|H) = (1/3) / (1/4 + 1/3)

P(B|H) = (1/3) / (3/12 + 4/12)

P(B|H) = (1/3) / (7/12)

P(B|H) = (1/3) * (12/7)

P(B|H) ≈ 0.571

So the probability that John picked the unbalanced coin given that he got heads is approximately 0.571 or 57.1%.

1/3 probability