# Thread: Probablity Question - have I done this right?

1. ## Probablity Question - have I done this right?

Hi there! I'm wondering if I've done the question below right ...

The teacher shows the students two bags. Bag A contains 3 gold coins and 2 silver coins. Bag B contains 4 gold coins and 5 silver coins. A coin is picked at random from Bag A and placed in Bag B. A coin is then picked at random from Bag B and placed in Bag A. These two operations are performed again in the same order. What is the probability that Bag A now contains five gold coins?

This is how I did it:
- In order to end up with 5 coins in A, each of the two swaps has to consist of putting gold coin into A from B and putting a silver coin into B from A.
- Probability of swaps occurring as described = probability of 1st swap x probability of 2nd swap = (4/9 x 2/5)(3/9 x 1/5) = 8/675

Does my logic follow? I'm curious to know as I've nowhere to check the answer!

2. ## Re: Probablity Question - have I done this right?

Originally Posted by HelenMc9
The teacher shows the students two bags. Bag A contains 3 gold coins and 2 silver coins. Bag B contains 4 gold coins and 5 silver coins. A coin is picked at random from Bag A and placed in Bag B. A coin is then picked at random from Bag B and placed in Bag A. These two operations are performed again in the same order. What is the probability that Bag A now contains five gold coins?

This is how I did it: In order to end up with 5 coins in A, each of the two swaps has to consist of putting gold coin into A from B and putting a silver coin into B from A.
- Probability of swaps occurring as described = probability of 1st swap x probability of 2nd swap = (4/9 x 2/5)(3/9 x 1/5) = 8/675
I have no idea where you are getting those numbers. Where do you get (4/9)?

In order to end with all gold coins in A we must gave the string $S_1G_2S_3G_4$.

That is, first ball is silver from A, second is gold from B, third is silver from A, and fourth is gold from B.

Find $\mathcal{P}(S_1G_2S_3G_4)=\mathcal{P}(S_1)\mathcal {P}(G_2|S_1)\mathcal{P}(S_3|S_1G_2)\mathcal{P}(G_4 |S_1G_2S_3)$