SOLUTION: Question 1 - Consider two urns. Urn 1 contains 4 red balls and 2 white balls, and urn 2 contains 3 balls of each color. If 2 balls are drawn from urn 1 without replacement and tran

Question 901356: Question 1 - Consider two urns. Urn 1 contains 4 red balls and 2 white balls, and urn 2 contains 3 balls of each color. If 2 balls are drawn from urn 1 without replacement and transferred to urn 2 and then a ball is drawn from urn 2, what is the probability that the ball drawn from urn 2 will be red . Given that the ball drawn from urn 2 was red, what is the conditional probability that (a) 0 (b) 1 (c) 2 red balls are transfered
Question 2- Let urn 1 contains 4 red balls and 2 white balls, and let the urn 2 contain 3 balls of each color. If a ball is drawn at random from urn 1 and transferred to urn 2 and then a ball is drawn at random from urn 2, what is the probability that the second ball drawn will be red
Found 2 solutions by solver91311, Natolino1983:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

If you ask two questions in your post when the rules for posting say "One question per post," what is the probability that you will get either one of them answered? Answer: Zero.

John

My calculator said it, I believe it, that settles it

Answer by Natolino1983(23) (Show Source): You can put this solution on YOUR website!
Q1) Case 1: 2 Red Balls are transfered from Urn1 to Urn 2.
Case 2: 1 Red Ball and 1 White Ball are transfered from Urn1 to Urn 2.
Case 3: 2 Red White Balls are transfered from Urn1 to Urn 2.
P(Case 1)= P (First Draw Red & Second Draw Red)=P(First Draw Red) *(Second Draw Red/First Draw Red) = 4/6 * 3/5 = 2/5.
P(Case 2)= P ((First Draw Red & Second Draw white) or (First draw White & Second draw Red))= P(First Draw red)*P(Second Draw White/First Draw Red) + P(First Draw White)*P(Second Draw Red/First Draw White) =4/6 * 2/5 + 2/6 * 4/5 = 4/15 + 4/15 = 8/15.
P(Case 3)= P (First Draw White & Second Draw White)=P(First Draw White) *(Second Draw White/First Draw White) = 2/6 * 1/5 = 1/15.
Note That P(Case 1)+ P(Case 2) + P(Case 3) = 1. (obviously).
So, 1)P(Draw Red Urn 2) = P(Draw Red Urn 2/Case 1) * (P(Case 1) + P(Draw Red Urn 2/Case 2) * (P(Case 2) + P(Draw Red Urn 2/Case 3) * (P(Case 3) = 5/8 * 2/5 + 4/8 * 8/15 + 3/8 * 1/15 = 1/4 + 4/16 + 1/40 = 21/40.
2)a)P(Case 3/Draw Red Urn 2) = (P(Draw Red Urn 2 /Case 3)* (P(Case 3))/ P(Draw Red Urn 2) = (1/40)/(21/40) =1/21
2 b) P(Case 2/Draw Red Urn 2) = (P(Draw Red Urn 2 /Case 2)* (P(Case 2))/ P(Draw Red Urn 2) = (4/16)/(21/40) =10/21.
2 c) P(Case 1/Draw Red Urn 2) = 1 - P(Case 3/Draw Red Urn 2) - P(Case 2/Draw Red Urn 2) = 1 - 1/21 -10/21 =10/21.
You can use Question 1 to solve question 2.

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