Finite math question

Question

Urn 1 contains 8 red and 2 white balls. Urn 2 contains 7 red and 2 white balls a ball is drawn from urn 1 and placed in urn 2 then a ball is drawn from urn 2 if the ball drawn from urn 2 is red what is the probability that the ball drawn from urn 1 was red

Answer 1

Urn 1 has 8 red balls and 2 white balls

Urn 2 has 7 red balls and 2 white balls

a ball is randomly taken from Urn 1 and put into Urn 2

then a ball is taken from Urn 2. If it's red, what is the probability the ball taken from Urn 1 was also red?? = about 84%

solve using Bayes Theorem with conditional probabilities

P(a/b)P(b) = P(b/a)P(a) where a=R and b = Urn1

P(R/Urn1)(P(Urn1) = P(Urn1/R)P(R)

P(R/Urn1) = P(Urn1/R)P(R)/P(Urn1) = (8/15)(15/19)/1/2) = 16(15)/15)(19) = 16/19 = about 84.21%

P(R/Urn1) = 8/10 = ..8 = 8/10 = 80%

P(R/Urn2) = 7/9 = about 78%

P(R) = 15/19

P(W = 4/19

find P(R/Urn1 if R/Urn2

Answer 2

Sample space:

 

R R = .8 * . 8 = .64

R W = .8 * . 2 = .16

W R = .2 * .7 = .14

W W = .2 * .3 = .06

 

Event A = red from urn 1....P(A) = .8

Event B = red from urn 2....p(B) = .78

P(A ∩ B) = .64  

P(A|B) = P(A ∩ B)/P(B) = .64/.78 = .820513