Question 119471: sirs,
please help me regarding this problem:
"there are 3 urns containing several balls. Urn 1 has 5 white, 3 green, and 2 red balls. Urn 2 contains 1 white, 6 green and 3 red balls. Urn 3 contains 3 white, 1 green and 6 red balls. a die is tossed once. if 1 appears, a ball is drawn from urn 1. if 2 or 3 appears, a ball is drawn from urn 2. otherwise, a ball is drawn from urn 3. determine (a) the probability that a red ball is chosen, and (b) the probability that urn 3 is chosen, given that the red ball is chosen."
thank you very much
sincerely,
cebrian
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! OK, let's see if we can simplify it.
Die 1, Urn 1
Die 2, Urn 2
Die 3, Urn 2
Die 4, Urn 3
Die 5, Urn 3
Die 6, Urn 3
Urn 1:5 W, 3 G, 2R=10 total
Urn 2:1 W, 6 G, 3R=10 total
Urn 3:3 W, 1 G, 6R=10 total
where W is white, G is green, and R is red.
a.) First, let's calculate the probability of a red ball for each urn.
P(R/U1)=2/10
P(R/U2)=3/10
P(R/U3)=6/10
P(die,1)=1/6
P(die,2)=1/6
P(die,3)=1/6
P(die,4)=1/6
P(die,5)=1/6
P(die,6)=1/6
P(R)=P(die,1)P(R/U1)+P(die,2)P(R/U2)+P(die,3)P(R/U2)+P(die,4)P(R/U3)+P(die,5)P(R/U3)+P(die,6)P(R/U3)
P(R)=1/6*2/10+1/6*3/10+1/6*3/10+1/6*6/10+1/6*6/10+1/6*6/10
P(R)=26/60=13/30
b)We're looking for the probability of U3 given the ball is red or P(U3/R).
Use Bayes' theorem.
P(U3/R)=P(R/U3)P(U3)/P(R)
P(R/U3)=6/10, from above
P(U3)=3/6, since there are 3 chances out of 6 from the die.
P(R)=13/30, since that was the answer to part a.
P(U3/R)=(6/10*3/6)/(13/30)
P(U3/R)=(6*3*30)/(6*10*13)
P(U3/R)=9/13
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