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?Options1/31/41/51/7
Question
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?Options1/31/41/51/7
Solution
To solve this problem, we first need to calculate the total number of bags in each room and the total number of green bags in each room.
In the first room, there are 3 blue + 4 green + 5 red = 12 bags in total, of which 4 are green.
In the second room, there are 2 blue + 1 green + 3 red = 6 bags in total, of which 1 is green.
The probability of the man picking a green bag from the first room is 4/12 = 1/3, and from the second room is 1/6.
Since the man picks a room at random, the probability of him picking any room is 1/2.
Therefore, the total probability of him picking a green bag is the sum of the probabilities of him picking a green bag from each room, each multiplied by the probability of him picking that room.
So, the total probability = (1/2 * 1/3) + (1/2 * 1/6) = 1/6 + 1/12 = 1/4.
So, the answer is 1/4.
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