Skip to content
Part BCSIR NET December 2024the-weights-cancel-against-the-white-fraction

The weights cancel against the white fraction

Let U1,U2,…,U5U_{1}, U_{2}, \dots, U_{5} be 5 urns such that urn U_k contains 2k+k22k + k^{2} balls, out of which 2k are white balls and k2k^{2} are black balls, k = 1, 2, …, 5. An urn is selected with probability of selecting urn U_k being proportional to (k + 2). A ball is chosen randomly from the selected urn. Then, the probability that the urn U5U_{5} was selected, given that the ball drawn is white, is equal to

  1. A.3/5
  2. B.2/5
  3. C.1/5✓ Correct
  4. D.3/4

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Pairwise independent events are mutually independent

More on this topic

The chapter behind this: Probability axioms, conditioning and independence — free to read

From Probability › Axioms, conditional probability, independence, Bayes

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram