Each of the `n` urns contains 4 white and 6 black balls. The `(n+1)` th urn contains 5 white and 5 black balls. One of the `n+1` urns is chosen at random and two balls are drawn from it without replacement. Both the balls turn out to be black. If the probability that the `(n+1)` th urn was chosen to draw the balls is 1/16, then find the value of `n` .

Explanation

To solve the problem, we will use Bayes' theorem and the concept of conditional probability. Let's break down the solution step by step.

### Step 1: Define Events
Let:
- \( E_1 \): The event that one of the first \( n \) urns is chosen.
- \( E_2 \): The event that the \( (n+1) \)th urn is chosen.
- \( A \): The event that both drawn balls are black.

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