Multiple choice

There are $\mathrm{n}$ urns each containing $\mathrm{n}+1$ balls such that the ith urn contains $\mathrm{i}$ white balls and $(\mathrm{n}+1-\mathrm{i})$ red balls. Let $\mathrm{u}{\mathrm{i}}$ be the event of selecting ith urn, $\mathrm{i}=1,2,3\ldots,\ \mathrm{n}$ and $\mathrm{w}$ denotes the event of getting a white ball. $\mathrm{I}\mathrm{f}\mathrm{P}(\mathrm{u}{\mathrm{i}})=\mathrm{c}$, where $\mathrm{c}$ is a constant then $\mathrm{P}(\mathrm{u}_{\mathrm{n}}/\mathrm{w})$ is equal to

  1. $\displaystyle \frac{2}{\mathrm{n}+1}$
  2. $\displaystyle \frac{1}{\mathrm{n}+1}$
  3. $\displaystyle \frac{\mathrm{n}}{\mathrm{n}+1}$
  4. $\displaystyle \frac{1}{2}$
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A Correct answer
Explanation

Using Bayes' theorem: P(Un/W) = P(W/Un) * P(Un) / sum(P(W/Ui) * P(Ui)). Since P(Ui) is constant, P(Un/W) = P(W/Un) / sum(P(W/Ui)). P(W/Ui) = i/(n+1). Sum = (1/(n+1)) * sum(i from 1 to n) = (1/(n+1)) * (n(n+1)/2) = n/2. P(W/Un) = n/(n+1). So P(Un/W) = (n/(n+1)) / (n/2) = 2/(n+1).

AI explanation

Using Bayes' theorem, P(un/w) equals P(un intersection w) divided by P(w). Since selecting an urn is equally likely, P(ui) equals 1/n, and the probability of drawing a white ball from urn n is n divided by (n+1). The total probability P(w) equals the sum from i equals 1 to n of (1/n) multiplied by (i divided by (n+1)), which evaluates to (n+1) divided by (2n). Substituting these values gives (1/n multiplied by n divided by (n+1)) divided by ((n+1) divided by (2n)), which simplifies to 2 divided by (n+1). Therefore, the result is 2 divided by (n+1).

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