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Each of the n urns contains 4 white and ...

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` .

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To solve the problem step by step, we will use the concepts of conditional probability and the law of total probability. ### Step 1: Define Events Let: - \( E_1 \): Event that the ball is chosen from one of the \( n \) urns. - \( E_2 \): Event that the ball is chosen from the \( (n+1) \)th urn. - \( A \): Event that two balls drawn are black. ...
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