Multiple choice

There are 4 urns. The first urn contains 1 white & 1 black ball, the second urn contains 2 white & 3 black balls, the third urn contains 3 white & 5 black balls & the fourth urn contains 4 white & 7 black balls. The selection of each urn is not equally likely. The probability of selecting $\displaystyle i^{th}$ urn is $\displaystyle \frac{i+1}{34}$ (i = 1, 2, 3, 4). If we randomly select one of the urn & draw a ball, then the probability of ball being white is

  1. $\displaystyle \frac{569}{1496}$
  2. $\displaystyle \frac{27}{56}$
  3. $\displaystyle \frac{8}{73}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
AI explanation

Using the law of total probability, we multiply the probability of selecting each urn by the probability of drawing a white ball from that urn, then sum the results. The weighted sum is calculated as (2/34)(1/2) + (3/34)(2/5) + (4/34)(3/8) + (5/34)(4/11). Evaluating this gives 1/34 + 3/85 + 3/34 + 10/74, which equals 569/1496. The result is 569/1496.

Keep practicing — related questions