Question

2 identical urns contain balls one of the urns has 6 red balls and 3 blue balls the other urn has 5 red balls and 8 blue balls and urn is chosen at random and 2 balls are drawn at random from this urn without replacement what is the probability that the second ball is red given that the first ball is red

Answer 1

Answer:

p(F∩S)=p((F∩S) ∩E1)+ p((F∩S) ∩E2)= p((F∩S) |E1)p(E1)+

p((F∩S)|E2)p(E2)=(6/9)(5/8)(1/2)+(5/13)(4/12)(1/2)=85/312

(b). We need p(S|F). By Bayes’ Theorem p(S|F)=p(S∩F)/p(F). p(S∩F) was

already computed in part a. We just need p(F).

P(F)=p(F∩E1)+p(F∩E2)=p(F|E1)p(E1)+p(F|E2)p(E2)=(6/9)(1/2)+(5/1

3)(1/2)

P(S|F)=0.5183