Urn I contains three red chips and five white chips; urn II contains four red sand four whites; urn III contains five red sand three whites. One urn is chosen at random and one chip is drawn from that urn. Given that the chip drawn was red, what is theprobability that III was the urn sampled?

Accepted Solution

Answer:0.416Step-by-step explanation:This problem is solve by using Baye's Probability.Let P(U1) = Probability that Urn I is selected = 1÷3P(U2) = Probability that Urn II is selected = 1÷3P(U3) = Probability that Urn III is selected = 1÷3Alos Let, R = Red ball is chosenand W = White ball is chosenThen, P(R/U1) = 3÷8P(W/U1) = 5÷8P(R/U2) = 4÷8P(W/U2) = 4÷8and, P(R/U3) = 5÷8P(W/U3) = 3÷8Then by Baye's Theorem,[tex]P(U3| R) = \dfrac{P(R/U3) \times P(U3)}{P(R/U1) \times P(U1) + P(R/U2) \times P(U2) + P(R/U3) \times P(U3)}[/tex]⇒  [tex]P(U3| R) = \dfrac{\frac{5}{8} \times\frac{1}{3} }{\frac{3}{8} \times\frac{1}{3}+\frac{4}{8} \times\frac{1}{3}+\frac{5}{8} \times\frac{1}{3}}[/tex]⇒ P(U3| R) = 0.416which is required probability.