Probability problem?

Question

One purse contains 6 copper and 1 gold coins. Another purse contains 4 copper coins. 5 coins are drawn from the first purse and put into the second purse. Then 2 coins are drawn from the second purse and put into the first purse. Determine the probability that the gold coin is in (1) 2nd purse (2) 1st purse. [2nd, 1st]
(A) 8/9, 4/9 (B) 7/9, 4/9 (C) 5/9, 2/9 (D) 5/9, 7/9 (E) 5/9, 4/9

I need to know how to go about solving this problem! help!!!

Answer 1

Probability for the gold coin to be in purse 1 = a + b

where a = probability that it moves to and forth
b= probability that it remains in purse 1 itself without being randomly selected.

Purse 1: 6 C and 1G coins...totally 7.
Lets calculate the probabilities a and b individually:

a) out of 7 coins 5 are chosen at random--> number of chances = 7C5 (I hope you understand the difficulty in typing it correctly)
out of these 5 coins, one is gold so remaining 4 are picked from the rest of the 6 coins = 6C4
probability that the gold coin moves into purse 2 is : 6C4/7C5

now there are 4+5 coins in purse 2 and two are chosen at random and the gold coin should get back to purse 1:
total number of chances = 9C2
chances of one gold coin to be among the 2 chosen coins : 8C1 or 8

now probability that the gold coin moves back to purse 1: 8/9C2

the product of these 2 probabilities is the one which we need to calculate: 6C4/7C5 * 8/9C2 = 10/63

b) probability that the gold coin remains in purse 1 itself (without moving)
7 coins; 5 chosen at random leaving the gold coin: 6C5/7C5= 2/7

so the over all probability = sum of individual probabilities: 10/63 + 2/7 = 4/9
probability that the coin is in purse 2 = 1-4/9 = 5/9

Answer 2

Logically, yljacktt is correct. But that doesn't show how it's done.

For the gold coin to be in the second purse, it has to be drawn the first time and not the second. For it to be in the first purse, either it has to be not drawn at all or drawn twice.

Not selected on first draw: 2/7 (in first purse)

Selected on first draw: 5/7
- Not selected on second draw: 7/9 (in second purse)
- Selected on second draw: 2/9 (in first purse)

Probability it is in first purse:
2/7 + 2/9 * 5/7 = 18/63 + 10/63 = 28/63 = 4/9

Probability it is in second purse:
7/9 * 5/7 = 5/9

Answer 3

Let

P(1+) - probability that gold coin remains in the 1st purse after 1st draw.
P(1-) - probability that gold coin is replaced after 1st draw.
P(2+) - probability that gold coin will be returned to the 1st purse (if it is in the 2nd purse) after 2nd draw.
P(2-) - probability that gold coin remains in the 2nd purse after 2nd draw.
P(1) - probability that the gold coin is finally in the 1st purse.
P(2) - probability that the gold coin is finally in the 2nd purse.

Please calculate the following probabilities:

P(1+) = ... (that's easy)
P(1-) = 1 - P(1+)
P(2+) = ... (that's easy too)
P(2-) = 1 - P(2+)

Now calculate the final result:

P(1) = P(1+) + P(1-)*P(2+)
P(2) = 1 - P(1)

Good luck.
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Answer 4

(2)♦ the probability the g-coin is still in #1 purse is p1=1/6;
♥ the probability the g-coin is 1 among 5 is p2=5/6, and the probability the g-coin is 1 of 2 put back p3=2/(4+5); q2=p1+p2*p3=19/54;
(1)♠ the probability the g-coin is not 1 of 2 put back p4=1-p3;
♣ q1 = p2*p4 = 35/54;
q1+q2=1 correct!

Answer 5

Well, think about this!
Wouldn't the probabilies add to 1?
So, which one adds up to 1?
(E) 5/9, 4/9.