A old man lives in a apartment containing 2 rooms. Each day before going to work he enters any one room randomly, picks up a bag and leaves home. One of the rooms contains 3 blue, 4 green and 5 red bags and the other contains 2 blue, 1 green and 3 red bags. What is the probability that he takes a green bag to work?

To solve this problem, we first need to calculate the total number of bags in each room and the total number of bags overall.

In the first room, there are 3 blue + 4 green + 5 red = 12 bags. In the second room, there are 2 blue + 1 green + 3 red = 6 bags. So, the total number of bags is 12 + 6 = 18 bags.

Next, we calculate the total number of green bags, which is 4 (from the first room) + 1 (from the second room) = 5 green bags.

The probability of the man picking a bag from the first room is the number of bags in the first room divided by the total number of bags, which is 12/18 = 2/3. Similarly, the probability of the man picking a bag from the second room is 6/18 = 1/3.

The probability of the man picking a green bag from the first room is the number of green bags in the first room divided by the total number of bags in the first room, which is 4/12 = 1/3. Similarly, the probability of the man picking a green bag from the second room is 1/6.

So, the total probability of the man picking a green bag is the sum of the probability of him picking a green bag from the first room and the probability of him picking a green bag from the second room.

This is calculated as (probability of picking from first room * probability of picking a green bag from first room) + (probability of picking from second room * probability of picking a green bag from second room).

So, the total probability = (2/3 * 1/3) + (1/3 * 1/6) = 2/9 + 1/18 = 4/18 + 1/18 = 5/18.