3 urns contain white and black balls. In the first – 2 white and 3 black balls, in the second

3 urns contain white and black balls. In the first – 2 white and 3 black balls, in the second – 2 white and 2 black balls, in the third – 3 white and one black balls. The ball was transferred from the first urn to the second. After that, the ball from the second urn was transferred to the third, Finally, from the third urn, the ball was transferred to the first. Determine the probability that the composition of the balls in all urns will remain unchanged.

So that the composition of the balls in the 3 urns does not change, it is necessary that each time only the white ball or only the black one is moved.

Let’s calculate the probabilities of this event and sum them up.

1) The probability of taking a white ball from the 1st urn = 2/5. After that, there will be 3 white balls in the 2nd urn.

The probability of catching a white ball from the 2nd urn = 3/5. After that, there will be 4 white balls in the 3rd urn.

The probability of taking a white ball from the third urn is 4/5.

We multiply the probabilities to take the white ball and get: 2/5 * 3/5 * 4/5 = 24/125.

2) Now let’s calculate the probability of choosing a black ball from the 1st urn. It is 3/5. After that, there will be 3 black balls in the 2nd urn, and the probability of getting it will be 3/5. Then there will be 2 black balls in the 3rd urn. The probability of getting a black ball from the 3rd urn will be 2/5.

Let’s multiply the probabilities of choosing a black ball from all urns and get: 3/5 * 3/5 * 2/5 = 18/125.

3) Overall probability = 24/125 + 18/125 = 42/125 = 0.336. Multiply by 100% and get 33.6%.

Answer: the probability that the balls will remain unchanged in all urns is 33.6%.