Multiple choice

Each of the urns contains $4$ white and $6$ blackballs. The (n + 1)th urn contains $5$ white and $5$black balls. One of the (n+1) urns is chosen atrandom and two balls are drawn from it withoutreplacement and both the balls turn out to beblack. Then the probability that the (n+1)th urn was chosen to draw the balls is $1/16$, the value of n is

  1. $10$
  2. $11$
  3. $12$
  4. $13$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let n urns have 4W, 6B and 1 urn have 5W, 5B. Total urns = n+1. P(Urn_n+1 | 2B) = [P(2B | Urn_n+1) * P(Urn_n+1)] / [P(2B | Urn_n+1) * P(Urn_n+1) + P(2B | Urn_n) * P(Urn_n)]. P(2B | Urn_n+1) = 5C2 / 10C2 = 10/45 = 2/9. P(2B | Urn_n) = 6C2 / 10C2 = 15/45 = 1/3. P(Urn_n+1) = 1/(n+1), P(Urn_n) = n/(n+1). 1/16 = (2/9 * 1/(n+1)) / (2/9 * 1/(n+1) + 1/3 * n/(n+1)) = (2/9) / (2/9 + n/3) = 2 / (2 + 3n). 2 + 3n = 32, 3n = 30, n = 10.

AI explanation

Using Bayes theorem, the conditional probability that the (n+1)th urn was chosen given two black balls are drawn is (1/(n+1)) * ((5C2)/(10C2)) divided by the total probability. The total probability of drawing two black balls is ((1/(n+1)) * ((n * (6C2/10C2)) + (5C2/10C2))), which simplifies to ((1/(n+1)) * ((n*(1/3)) + (2/9))). Setting the conditional probability ratio equal to 1/16 gives (2/9) / ((n/3) + (2/9)) = 1/16. Solving this equation, 32 = 3n + 2, results in 3n = 30, so n is 10.

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