- 2 82 An Urn Contains 11 Amber Balls And 17 Black Balls Two Balls Are Selected At Random Without Replacement And Remov 1 (69.88 KiB) Viewed 66 times
- 2 82 An Urn Contains 11 Amber Balls And 17 Black Balls Two Balls Are Selected At Random Without Replacement And Remov 2 (201.1 KiB) Viewed 66 times
2.82 An urn contains 11 amber balls and 17 black balls. Two balls are selected at random, without replacement, and removed from the urn. The colors of the two balls removed are not observed. A third ball is then drawn from the urn at random. Find the probability that the third ball removed from the urn is amber.
2.82 A, A; A₂ = removing amberball on draw i و اع 11 = B 17 B., B2 2 black ball PLA) = P(A3|8, 08.) PCB NB) ) PABB₂ P ( + 7 (А, В, ПА) Р/, ПА) + PLASIA, NB) PLA, 2B) + PLAz 1 A, A) PLANA) Note: Now Answer ) mult rule Lot loend