There are 3 true coins and 1 false coin with ‘head’ on both sides.A coin is chosen at random and tossed 4 times. If ‘head’ occurs allthe four times, what is the probability that the false coin has beenchosen and used?

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

A: The false coin is chosen and used. B: 'Head' occurs all four times.

We want to find the probability of event A given event B, P(A|B).

According to Bayes' theorem, we have:

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

P(B|A) is the probability of getting 'head' all four times given that the false coin is chosen. Since the false coin has 'head' on both sides, the probability of getting 'head' in a single toss is 1. Therefore, the probability of getting 'head' all four times given that the false coin is chosen is (1^4) = 1.

P(A) is the probability of choosing the false coin. Out of the 4 coins, there is only 1 false coin. Therefore, P(A) = 1/4.

P(B) is the probability of getting 'head' all four times. We can calculate this by considering all possible scenarios:

1. Choosing the false coin and getting 'head' all four times: (1/4) * 1 = 1/4.
2. Choosing a true coin and getting 'head' all four times: (3/4) * (1/2)^4 = 3/32.

Therefore, P(B) = (1/4) + (3/32) = 11/32.

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

P(A|B) = (1 * 1/4) / (11/32) = 8/11.

So, the probability that the false coin has been chosen and used, given that 'head' occurs all four times, is 8/11.